GET 20 SSC GD MOCK FOR JUST RS. 117<\/a><\/p>\n FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n Download SSC GD Important Questions PDF<\/a><\/p>\n 1500<\/strong>+ Must Solve Questions for SSC Exams<\/a> (Question bank)<\/p>\n Question 1:\u00a0<\/b>What is the equation of a line which passes through the point of intersection of lines A:x-3y=7 and B:2x+y=-7 and the required line is also perpendicular to line x+4y=10 ?<\/p>\n a)\u00a04x-y+4=0<\/p>\n b)\u00a04x-y+5=0<\/p>\n c)\u00a04x-3y-1=0<\/p>\n d)\u00a0x-4y-10=0<\/p>\n Question 2:\u00a0<\/b>Equation of one of the diagonals of a rhombus is 2x+3y=10. what is the equation of the other diagonal passing through the point (1,3) ?<\/p>\n a)\u00a0$3x+2y-9=0$<\/p>\n b)\u00a0$6x-4y+5=0$<\/p>\n c)\u00a0$3x-2y+3=0$<\/p>\n d)\u00a0$2x+3y-11=0$<\/p>\n Question 3:\u00a0<\/b>The straight line y=3x must pass through the point:<\/p>\n a)\u00a0(0,1)<\/p>\n b)\u00a0(0,0)<\/p>\n c)\u00a0(2,0)<\/p>\n d)\u00a0(1,2)<\/p>\n Question 4:\u00a0<\/b>A is a point on the x-axis with abscissa – 8 and B is the pint on the y-axis with ordinate 15. The length of AB is<\/p>\n a)\u00a019 units<\/p>\n b)\u00a021 units<\/p>\n c)\u00a017 units<\/p>\n d)\u00a023 units<\/p>\n Question 5:\u00a0<\/b>What is the reflection of the point (-5, -2) in the line y = 1?<\/p>\n a)\u00a0(-5, 1)<\/p>\n b)\u00a0(-5, 4)<\/p>\n c)\u00a0(-5, 0)<\/p>\n d)\u00a0(-5, 5)<\/p>\n Download SSC GD FREE PREVIOUS PAPERS<\/a><\/p>\n LATEST FREE SSC GD MOCK 2019<\/a><\/p>\n Question 6:\u00a0<\/b>If (1, 3) is the centroid of triangle ABC, with co-ordinates A(1, 6), B(3 , a) & C(b, 1), what is the value of a + b?<\/p>\n a)\u00a05<\/p>\n b)\u00a01<\/p>\n c)\u00a06<\/p>\n d)\u00a04<\/p>\n Question 7:\u00a0<\/b>What is the reflection of the point ( 6, -3) about the line y = -3?<\/p>\n a)\u00a0(6, 3)<\/p>\n b)\u00a0(6, 0)<\/p>\n c)\u00a0(6, -6)<\/p>\n d)\u00a0(6, -3)<\/p>\n Question 8:\u00a0<\/b>Find the area (in sq units) enclosed by y = 0, x+6=0 and 2x-3y=6.<\/p>\n a)\u00a09<\/p>\n b)\u00a016<\/p>\n c)\u00a020<\/p>\n d)\u00a027<\/p>\n Question 9:\u00a0<\/b>What is the area of the triangle made by line 3x+4y = 24 with the coordinate axes?<\/p>\n a)\u00a06 square units<\/p>\n b)\u00a012 square units<\/p>\n c)\u00a024 square units<\/p>\n d)\u00a048 square units<\/p>\n Question 10:\u00a0<\/b>A cube is cut into 8 equal smaller cubes.The percentage change in the surface area is<\/p>\n a)\u00a080%<\/p>\n b)\u00a050%<\/p>\n c)\u00a0150%<\/p>\n d)\u00a0100%<\/p>\n DOWNLOAD APP TO ACESS DIRECTLY ON MOBILE<\/a><\/p>\n Daily Free SSC Practice Set<\/a><\/p>\n Question 11:\u00a0<\/b>Find the centroid of the triangle formed by the vertices, P(-4,-2) Q(-3, -5) and R(1, 1) ?<\/p>\n a)\u00a0(2, -2)<\/p>\n b)\u00a0(-2, 2)<\/p>\n c)\u00a0(-2,-2)<\/p>\n d)\u00a0(2, 2)<\/p>\n Question 12:\u00a0<\/b>Which of the following represents hyperbola ?<\/p>\n a)\u00a0$x^2+y^2 = a^2$<\/p>\n b)\u00a0$\\frac{x^2}{a^2}+\\frac{y^2}{b^2} = 1$<\/p>\n c)\u00a0$\\frac{x^2}{a^2}-\\frac{y^2}{b^2} = 1$<\/p>\n d)\u00a0None of the above<\/p>\n Question 13:\u00a0<\/b>Find the area of triangle formed by the line 6x+7y = 21 with coordinate axes ?<\/p>\n a)\u00a05.25 sq.units<\/p>\n b)\u00a06.25 sq.units<\/p>\n c)\u00a09.25 sq.units<\/p>\n d)\u00a012.25 sq.units<\/p>\n Question 14:\u00a0<\/b>What is the reflection of the point ( -2, -3) in the line y = -4?<\/p>\n a)\u00a0(-2, -2)<\/p>\n b)\u00a0(-2, -5)<\/p>\n c)\u00a0(1, -2)<\/p>\n d)\u00a0(2, 5)<\/p>\n Question 15:\u00a0<\/b>What is the equation of a circle of radius 1 unit centered at (-5, -4) ?<\/p>\n a)\u00a0$x^2+y^2+10x+8y+40 = 0$<\/p>\n b)\u00a0$x^2+y^2-10x+8y+40 = 0$<\/p>\n c)\u00a0$x^2+y^2+10x-8y+40 = 0$<\/p>\n d)\u00a0$x^2+y^2+10x+8y-40 = 0$<\/p>\n SSC CGL Free Mock Test<\/a><\/p>\n Download SSC GD Constable Syllabus PDF<\/a><\/p>\n Question 16:\u00a0<\/b>If the ordinate and abscissa of (2k+3, 4+k) are equal, then find the value of k ?<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a03<\/p>\n Question 17:\u00a0<\/b>The coordinates of the vertices of a right angled triangle are A(4,3), B(8,-2) and C(4, -2). It is known that the triangle is right angled at C. Find the orthocenter of the triangle ABC ?<\/p>\n a)\u00a0(4,-2)<\/p>\n b)\u00a0(8,3)<\/p>\n c)\u00a0(6,3)<\/p>\n d)\u00a0(2,3)<\/p>\n Question 18:\u00a0<\/b>What is the reflection of the point ( -4, -4) in the line x = 1 ?<\/p>\n a)\u00a0(-6, 4)<\/p>\n b)\u00a0(-6, -4)<\/p>\n c)\u00a0(6, 4)<\/p>\n d)\u00a0(6, -4)<\/p>\n Question 19:\u00a0<\/b>Which of the following represents Ellipse ?<\/p>\n a)\u00a0$x^2-y^2 = a^2$<\/p>\n b)\u00a0$\\frac{x^2}{a^2}+\\frac{y^2}{b^2} = 1$<\/p>\n c)\u00a0$\\frac{x^2}{a^2}-\\frac{y^2}{b^2} = 1$<\/p>\n d)\u00a0None of the above<\/p>\n Question 20:\u00a0<\/b>Find the centroid of the triangle formed by the vertices, P(2,3) Q(-3, -5) and R(1, -4) ?<\/p>\n a)\u00a0(-1, 2)<\/p>\n b)\u00a0(0, 2)<\/p>\n c)\u00a0(1,-2)<\/p>\n d)\u00a0(0, -2)<\/p>\n Question 21:\u00a0<\/b>Find the area of triangle formed by the line 3x+5y = 15 with coordinate axes ?<\/p>\n a)\u00a03.5 sq.units<\/p>\n b)\u00a05.5 sq.units<\/p>\n c)\u00a07.5 sq.units<\/p>\n d)\u00a09.5 sq.units<\/p>\n Question 22:\u00a0<\/b>What is the reflection of the point ( -3, -2) in the line y = 0?<\/p>\n a)\u00a0(-3, 2)<\/p>\n b)\u00a0(3, 2)<\/p>\n c)\u00a0(3, -2)<\/p>\n d)\u00a0(-3, -2)<\/p>\n Question 23:\u00a0<\/b>If the ordinate and abscissa of (k+1, 1-k) are equal, then find the value of k ?<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a03<\/p>\n Question 24:\u00a0<\/b>What is the equation of a circle of radius 5 units centered at (-4, 2) ?<\/p>\n a)\u00a0$x^2+y^2-2x-4y+5=0$<\/p>\n b)\u00a0$x^2+y^2+8x-4y+5 = 0$<\/p>\n c)\u00a0$x^2+y^2-8x-4y-5 = 0$<\/p>\n d)\u00a0$x^2+y^2+8x-4y-5 = 0$<\/p>\n Question 25:\u00a0<\/b>The coordinates of the vertices of a right angled triangle are A(8,-2) B(8,3) and C(2, 3). It is known that the triangle is right angled at B. Find the orthocenter of the triangle ABC ?<\/p>\n a)\u00a0(2,-2)<\/p>\n b)\u00a0(8,3)<\/p>\n c)\u00a0(6,3)<\/p>\n d)\u00a0(2,3)<\/p>\n Question 26:\u00a0<\/b>What is the reflection of the point ( -2, -2) in the line x = 2?<\/p>\n a)\u00a0(-6, -2)<\/p>\n b)\u00a0(6, -2)<\/p>\n c)\u00a0(-6, 2)<\/p>\n d)\u00a0(6, 2)<\/p>\n 100+ Free GK Tests for SSC Exams<\/a><\/p>\n Download Free GK PDF<\/a><\/p>\n Question 27:\u00a0<\/b>Find the distance between the points P(2,4) and Q(-2,1) ?<\/p>\n a)\u00a04 units<\/p>\n b)\u00a05 units<\/p>\n c)\u00a06<\/p>\n d)\u00a07 units<\/p>\n Question 28:\u00a0<\/b>A right angled triangle A(2,3), B(8,3) and C(2, -1) is right angled at A. What is the orthocentre of triangle ABC ?<\/p>\n a)\u00a0(10,2)<\/p>\n b)\u00a0(5,2)<\/p>\n c)\u00a0(5,1)<\/p>\n d)\u00a0(2,3)<\/p>\n Question 29:\u00a0<\/b>Find the area of triangle formed by the line 2x+3y = 6 with coordinate axes ?<\/p>\n a)\u00a03 sq.units<\/p>\n b)\u00a06 sq.units<\/p>\n c)\u00a09 sq.units<\/p>\n d)\u00a012 sq.units<\/p>\n Question 30:\u00a0<\/b>What is the reflection of the point ( -1, 4) in the line y = -3?<\/p>\n a)\u00a0(1, -10)<\/p>\n b)\u00a0(-1, 10)<\/p>\n c)\u00a0(-1, -10)<\/p>\n d)\u00a0(1, 10)<\/p>\n Question 31:\u00a0<\/b>Find the slope of the line 3x-4y+4 = 0 ?<\/p>\n a)\u00a04\/3<\/p>\n b)\u00a03\/2<\/p>\n c)\u00a02\/3<\/p>\n d)\u00a03\/4<\/p>\n Question 32:\u00a0<\/b>What is the equation of a circle of radius 3 units centered at (-1, 3) ?<\/p>\n a)\u00a0$x^2+y^2-2x-4y+5=0$<\/p>\n b)\u00a0$x^2+y^2-2x-6y+1 = 0$<\/p>\n c)\u00a0$x^2+y^2+2x-6y+1 = 0$<\/p>\n d)\u00a0$x^2+y^2+2x-6y-1 = 0$<\/p>\n Question 33:\u00a0<\/b>If the ordinate and abscissa of (2k+3, 6-k) are equal, then find the value of k ?<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a04<\/p>\n Question 34:\u00a0<\/b>What is the reflection of the point ( -3, -2) in the line x = -1?<\/p>\n a)\u00a0(-1, -2)<\/p>\n b)\u00a0(-1, 2)<\/p>\n c)\u00a0(1, -2)<\/p>\n d)\u00a0(1, 2)<\/p>\n Question 35:\u00a0<\/b>The coordinates of the vertices of a right angled triangle are A(6,-7) B(8,3) and C(6, 3), which is right angled at C. Find the orthocenter of the triangle ABC ?<\/p>\n a)\u00a0(2,-2)<\/p>\n b)\u00a0(8,3)<\/p>\n c)\u00a0(6,3)<\/p>\n d)\u00a0(2,3)<\/p>\n Question 36:\u00a0<\/b>Find the value of k (greater than 0) for which the lines (2k-1)x+3y-2 = 0 and 3x+(4k-5)y+4 = 0 are parallel ?<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a04<\/p>\n Question 37:\u00a0<\/b>Find the area of a right angled triangle formed by the vertices A(6,-7) B(8,3) and C(6, 3), which is right angled at C ?<\/p>\n a)\u00a025 sq.units<\/p>\n b)\u00a020 sq.units<\/p>\n c)\u00a015 sq.units<\/p>\n d)\u00a010 sq.units<\/p>\n Question 38:\u00a0<\/b>Find the area of triangle formed by the line 4x+3y = 12 with coordinate axes ?<\/p>\n a)\u00a03 sq.units<\/p>\n b)\u00a06 sq.units<\/p>\n c)\u00a09 sq.units<\/p>\n d)\u00a012 sq.units<\/p>\n Question 39:\u00a0<\/b>What is the reflection of the point ( -2, 2) in the line y = -2?<\/p>\n a)\u00a0(-2, -6)<\/p>\n b)\u00a0(2, -6)<\/p>\n c)\u00a0(6, -2)<\/p>\n d)\u00a0(6, 2)<\/p>\n Question 40:\u00a0<\/b>If the ordinate and abscissa of (2k, 6-k) are equal, then find the value of k ?<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a04<\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The point of intersection of the lines A and B is (-2,-3) 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Diagonals of a rhombus are perpendicular to each other. 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Equation of line = $y=3x$<\/p>\n (A) : (0,1)<\/p>\n => L.H.S. = $1\\neq$ R.H.S. = $3(0)=0$<\/p>\n (B) : (0,0)<\/p>\n => L.H.S. = $0=$ R.H.S. = $3(0)=0$<\/p>\n => Ans – (B)<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Now, OA = 8 units and OB = 15 units<\/p>\n => $(AB)^2=(OA)^2+(OB)^2$<\/p>\n => $(AB)^2=(8)^2+(15)^2$<\/p>\n => $(AB)^2=64+225=289$<\/p>\n => $AB=\\sqrt{289}=17$ units<\/p>\n => Ans – (C)<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n We can see that (-5, -2) is 3 units away from y = 1 SSC GD FREE STUDY MATERIAL<\/a><\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Centroid = $(\\frac{x_{1}+x_{2}+x_{3}}{3} , \\frac{y_{1}+y_{2}+y_{3}}{3})$ $\\Rightarrow$ $1 = \\frac{1 + 3 + b}{3}$ Also, we know that $3 = \\frac{6 + a + 1}{3}$ Therefore, a+b = 2 -1 = 1. 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Since the given point (6, -3) lies on the line y = -3 itself, the point and its reflection will coincide. 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The given line will cut the x-axis at a point where the y co-ordinate will be equal to zero.<\/p>\n $\\Rightarrow$ 3x + 4*0 = 24.<\/p>\n $\\Rightarrow$ x = 8.<\/p>\n So, the point of intersection on the x axis will be (8, 0).<\/p>\n Similarly, the line will cut y-axis at a point where the x co-ordinate will be equal to zero.<\/p>\n $\\Rightarrow$ 3*0 + 4y = 24.<\/p>\n $\\Rightarrow$ y = 6.<\/p>\n So, the point of interaction on the y axis will be (0, 6)<\/p>\n We can see that $\\triangleOAB$ is right-angled at O.<\/p>\n So, the area of $\\triangleOAB$ = $\\frac{1}{2} * OA * OB$<\/p>\n = $\\frac{1}{2} * 6 * 8$ 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let us assume the side of the original cube to be 2a units. 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Centroid = ($\\frac{x_1+x_2+x_3}{3}, \\frac{y_1+y_2+y_3}{3}$) = ($\\frac{-4-3+1}{3}, \\frac{-2-5+1}{3}$) = (-2, -2)<\/p>\n So the answer is option C.<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Option A & B, represent Ellipse. ($\\because$ general form of Ellipse is $\\frac{x^2}{a^2}+\\frac{y^2}{b^2} = 1$)<\/p>\n Option C represents Hyperbola. ($\\because$ general form of Hyperbola is $\\frac{x^2}{a^2}-\\frac{y^2}{b^2} = 1$)<\/p>\n So the answer is option C.<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Area of the line $\\frac{x}{a}+\\frac{y}{b} = 1$, made by Coordinate axes is $\\frac{1}{2}ab$.<\/p>\n $6x+7y = 21 ==> \\frac{x}{3.5}+\\frac{y}{3} = 1$<\/p>\n Area = $\\frac{1}{2}ab = \\frac{1}{2}(3.5)(3) = 5.25 sq.units$<\/p>\n So the answer is option A.<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n (-2, -3) lies above the line y=-4<\/p>\n Distance b\/w the line & point is -3-(-4) = 1<\/p>\n So its reflection must be 1 unit below the line , i.e; (-2, -4-1) = (-2, -5)<\/p>\n So the answer is option B.<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Eq.of a circle of radius r and having centre at ($x_1, y_1$) is $(x-x_1)^2+(y-y_1)^2=r^2$<\/p>\n Hence, $(x+5)^2+(y+4)^2=1^2$<\/p>\n ==> $x^2+10x+25+y^2+8y+16 = 1$<\/p>\n ==> $x^2+y^2+10x+8y+40 = 0$<\/p>\n So the answer is option A.<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 2k+3 = 4+k<\/p>\n 2k-k = 4-3<\/p>\n k = 1<\/p>\n So the answer is option B.<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Orthocentre of a right angled triangle is the vertex where the angle is 90\u00b0<\/p>\n Here, it is right angled at C, so C(4, -2) is the orthocentre.<\/p>\n So the answer is option A.<\/p>\n 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n (-4, -4) lies on left to the line x=1<\/p>\n Distance b\/w the line & point is 1-(-4) = 5<\/p>\n So its reflection must be 5 units to the right side the line , i.e; (1+5, -4) = (6, -4)<\/p>\n So the answer is option D.<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Option A & C, represent hyperbola. ($\\because$ general form of hyperbola is $\\frac{x^2}{a^2}-\\frac{y^2}{b^2} = 1$)<\/p>\n Option B represents Ellipse. ($\\because$ general form of Ellipse is $\\frac{x^2}{a^2}+\\frac{y^2}{b^2} = 1$)<\/p>\n So the answer is option B.<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Centroid = ($\\frac{x_1+x_2+x_3}{3}, \\frac{y_1+y_2+y_3}{3}$) = ($\\frac{2-3+1}{3}, \\frac{3-5-4}{3}$) = (0, -2)<\/p>\n So the answer is option D.<\/p>\n 21)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Area of the line $\\frac{x}{a}+\\frac{y}{b} = 1$, made by Coordinate axes is $\\frac{1}{2}ab$.<\/p>\n $3x+5y = 15 ==> \\frac{x}{5}+\\frac{y}{3} = 1$<\/p>\n Area = $\\frac{1}{2}ab = \\frac{1}{2}(3)(5) = 7.5 sq.units$<\/p>\n So the answer is option C.<\/p>\n 22)\u00a0Answer\u00a0(A)<\/strong><\/p>\n (-3, -2) lies below the line y=0<\/p>\n Distance between the line & point is 0-(-2) = 2<\/p>\n So its reflection must be 2 units above the line , i.e; (-3, 0+2) = (-3, 2)<\/p>\n So the answer is option A.<\/p>\n 23)\u00a0Answer\u00a0(A)<\/strong><\/p>\n k+1 = 1-k<\/p>\n 2k = 0<\/p>\n k = 0<\/p>\n So the answer is option A.<\/p>\n 24)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Eq.of a circle of radius r and having centre at ($x_1, y_1$) is $(x-x_1)^2+(y-y_1)^2=r^2$<\/p>\n Hence, $(x+4)^2+(y-2)^2=5^2$<\/p>\n ==> $x^2+8x+16+y^2-4y+4 = 25$<\/p>\n ==> $x^2+y^2+8x-4y-5 = 0$<\/p>\n So the answer is option D.<\/p>\n 25)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Orthocentre of a right angled triangle is the vertex where the angle is 90\u00b0<\/p>\n Here, it is right angled at B, so B(8, 3) is the orthocentre.<\/p>\n So the answer is option B.<\/p>\n 26)\u00a0Answer\u00a0(B)<\/strong><\/p>\n (-2, -2) lies on left to the line x=2<\/p>\n Distance b\/w the line & point is 2-(-2) = 4<\/p>\n So its reflection must be 4 units to the right side the line , i.e; (2+4, -2) = (6, -2)<\/p>\n So the answer is option B.<\/p>\n 27)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Distance = $\\sqrt{(y_2-y_1)^2+(x_2-x_1)^2} = \\sqrt{(1-4)^2+(-2-2)^2} = \\sqrt{9+16} = 5$<\/p>\n So the answer is option B.<\/p>\n 28)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Orthocentre of a right angled triangle is the vertex where the angle is 90\u00b0<\/p>\n Here, it is right angled at A, so C(2, 3) is the Orthocentre.<\/p>\n So the answer is option D.<\/p>\n 29)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Area of the line $\\frac{x}{a}+\\frac{y}{b} = 1$, made by Coordinate axes is $\\frac{1}{2}ab$.<\/p>\n $2x+3y = 6 ==> \\frac{x}{3}+\\frac{y}{2} = 1$<\/p>\n Area = $\\frac{1}{2}ab = \\frac{1}{2}(3)(2) = 3 sq.units$<\/p>\n So the answer is option A.<\/p>\n 30)\u00a0Answer\u00a0(C)<\/strong><\/p>\n (-1,4) lies above the line y=-3<\/p>\n Distance b\/w the line & point is 4-(-3) = 7<\/p>\n So its reflection must be 7 units below the line , i.e; (-1, -3-7) = (-1, -10)<\/p>\n So the answer is option C.<\/p>\n 31)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Slope of the line ax+by+c = 0 is -a\/b<\/p>\n Hence slope of the line 3x-4y+4 = 0 is -3\/-4 = 3\/4<\/p>\n So the answer is option D.<\/p>\n 32)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Eq.of a circle of radius r and having centre at ($x_1, y_1$) is $(x-x_1)^2+(y-y_1)^2=r^2$<\/p>\n Hence, $(x+1)^2+(y-3)^2=3^2$<\/p>\n ==> $x^2+2x+1+y^2-6y+9 = 9$<\/p>\n ==> $x^2+y^2+2x-6y+1 = 0$<\/p>\n So the answer is option C.<\/p>\n 33)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 2k+3 = 6-k<\/p>\n 2k+k = 6-3<\/p>\n 3k = 3<\/p>\n k = 1<\/p>\n So the answer is option A.<\/p>\n 34)\u00a0Answer\u00a0(C)<\/strong><\/p>\n (-3, -2) lies on left to the line x=-1<\/p>\n Distance b\/w the line & point is -1-(-3) = 2<\/p>\n So its reflection must be 2 units to the right side the line , i.e; (-1+2, -2) = (1, -2)<\/p>\n So the answer is option C.<\/p>\n 35)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Orthocentre of a right angled triangle is the vertex where the angle is 90\u00b0<\/p>\n Here, it is right angled at C, so C(6, 3) is the orthocentre.<\/p>\n So the answer is option C.<\/p>\n 36)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Slope of the line ax+by+c = 0 is -a\/b<\/p>\n For parallel lines, slopes are equal<\/p>\n ==>$\\frac{-(2k-1)}{3} = \\frac{-3}{4k-5}$<\/p>\n ==>$8k^2-14k-4 = 0$<\/p>\n ==>$4k^2-7k-2 = 0$<\/p>\n ==>$(4k+1)(k-2) = 0$<\/p>\n ==>$k = -1\/4$ or $k = 2$<\/p>\n So the answer is option B.<\/p>\n 37)\u00a0Answer\u00a0(D)<\/strong><\/p>\n AC = 3-(-7) = 10<\/p>\n BC = 8-6 = 2<\/p>\n Area of given triangle = $\\frac{1}{2}(AC)(BC) = \\frac{1}{2}(10)(2) = 10 sq.cm$<\/p>\n So the answer is option D.<\/p>\n 38)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Area of the line $\\frac{x}{a}+\\frac{y}{b} = 1$, made by Coordinate axes is $\\frac{1}{2}ab$.<\/p>\n $4x+3y = 12 ==> \\frac{x}{3}+\\frac{y}{4} = 1$<\/p>\n Area = $\\frac{1}{2}ab = \\frac{1}{2}(3)(4) = 6 \\text{ sq.units}$<\/p>\n So the answer is option B.<\/p>\n 39)\u00a0Answer\u00a0(A)<\/strong><\/p>\n (-2, 2) lies above the line y=-2<\/p>\n Distance between the line & point is 2-(-2) = 4<\/p>\n So its reflection must be 4 units below the line , i.e; (-2, -2-4) = (-2, -6)<\/p>\n So the answer is option A.<\/p>\n 40)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 2k = 6-k<\/p>\n 3k = 6<\/p>\n k = 2<\/p>\n So the answer is option B.<\/p>\n
\nthe line is perpendicular to x+4y=10
\nSlope of x+4y=10 is ($\\frac{-1}{4}$)
\n$\\frac{-1}{4}\\times s $=-1
\ns=4
\nEquation of line is 4(x+2)=y+3
\n4x+8=y+3
\n4x-y+5=0<\/p>\n
\nLines perpendicular to each other have the product of their slopes = -1
\nSlope of the line 2x+3y-10=0 is $-(\\frac{a}{b})$
\nSlope = $\\frac{-2}{3}$
\nThen slope of the other line is $ \\frac{3}{2}$.
\n$\\therefore$ required equation is $\\frac{3}{2}(x-1)=(y-3)$
\n$3x-3=2y-6$
\n$ 3x-2y+3=0$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/31926.PNG\")
<\/figure>\n
\nX coordinate of the point will remain unchanged while y axis will change
\n$\\Rightarrow$ $1 = \\frac{-2 + b}{2}$
\n$\\Rightarrow$ $2 = -2 + b$
\n$\\Rightarrow$ $b = 4$
\nAlternatively from graph we can see the reflection of point as well. (Answer :B)<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Capture_302n0eD.PNG\")
<\/figure>\n
\n$(1, 3)$ = $(\\frac{1 + 3 + b}{3} , \\frac{6 + a + 1}{3})$<\/p>\n
\n$\\Rightarrow$ $b + 4 = 3$
\n$\\Rightarrow$ $b = -1$<\/p>\n
\n$\\Rightarrow$ $a + 7 = 9$
\n$\\Rightarrow$ $a = 2$<\/p>\n
\nHence, option B is the right answer.<\/p>\n
\nTherefore, the reflection of point = (6, -3).
\nHence, option D is the right answer.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/tRAINGLE.PNG\")
<\/figure>\n
\n= 24 sq units.
\nTherefore, option C is the right answer.<\/p>\n
\n=> The surface area of the bigger cube = $6*(2a)^{2} = 24(a)^{2}$
\nWhen we cut the original cube to get 8 smaller cubes, each side will become half of its original length.
\nSo, the side of smaller cube will be 2a\/2 = a units in length.
\nSurface area of 1 smaller cube = $6*(a)^{2} = 6(a)^{2}$
\nTotal surface area of 8 cubes combined = $8 * 6(a)^{2} = 48(a)^{2}$
\nSo, the percentage change in the surface area = $\\frac{48(a)^{2} – 24(a)^{2}}{24(a)^{2}} \\times 100$ = 100 %.
\nTherefore, option D is the right answer.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/118522_Ejjzqi2.png\")
<\/figure>\n