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>The coordinates of the vertices of a right angled triangle are A(2,3) B(8,3) and C(2, -2). The triangle is right angled at A. 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 2:\u00a0<\/b>What is the equation of a circle of radius 4 units centered at (1, 2) ?<\/p>\n a)\u00a0$x^2+y^2-2x-4y+5=0$<\/p>\n b)\u00a0 $x^2+y^2-2x-4y-11 = 0$<\/p>\n c)\u00a0 $x^2+y^2+2x-4y-11 = 0$<\/p>\n d)\u00a0 $x^2+y^2-2x+4y-11 = 0$<\/p>\n Question 3:\u00a0<\/b>What is the reflection of the point ( -4, 2) in the line x = -3?<\/p>\n a)\u00a0(-2, -2)<\/p>\n b)\u00a0(-2, 2)<\/p>\n c)\u00a0(1, -2)<\/p>\n d)\u00a0(1, 2)<\/p>\n Question 4:\u00a0<\/b>Find the slope of the line joining A(5,-1) & B(2, 2) ?<\/p>\n a)\u00a01<\/p>\n b)\u00a0-1<\/p>\n c)\u00a02<\/p>\n d)\u00a0-2<\/p>\n Question 5:\u00a0<\/b>Point P is the midpoint of segment AB. Coordinates of P are (-1 , -1) and B are (-5, -4) What are the coordinates of point A?<\/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 Download SSC GD FREE PREVIOUS PAPERS<\/a><\/p>\n LATEST FREE SSC GD MOCK 2019<\/a><\/p>\n Question 6:\u00a0<\/b>In what ratio does the point T(x,3) divide the segment joining the points S(3, -1) and U(-1, 2)?<\/p>\n a)\u00a04:1<\/p>\n b)\u00a0-4:1<\/p>\n c)\u00a01:4<\/p>\n d)\u00a0-1:4<\/p>\n Question 7:\u00a0<\/b>What is the reflection of the point (-4 , 3) in the line x = -1?<\/p>\n a)\u00a0(-2 , 3)<\/p>\n b)\u00a0(2 , -3)<\/p>\n c)\u00a0(-2 , -3)<\/p>\n d)\u00a0(2 , 3)<\/p>\n Question 8:\u00a0<\/b>At what point does the line 3x – 2y = 12 intersect the Y-axis?<\/p>\n a)\u00a0(0,-6)<\/p>\n b)\u00a0(-6,0)<\/p>\n c)\u00a0(0,6)<\/p>\n d)\u00a0(6,0)<\/p>\n Question 9:\u00a0<\/b>Find the slope of the line parallel to x-3y+2=0 ?<\/p>\n a)\u00a0-1\/3<\/p>\n b)\u00a01\/3<\/p>\n c)\u00a03<\/p>\n d)\u00a0-3<\/p>\n Question 10:\u00a0<\/b>Find the point which divides the line joining A(2,3) B(4,1) in the ratio 2:3 internally?<\/p>\n a)\u00a0(-14\/5, -11\/5)<\/p>\n b)\u00a0(14\/5, -11\/5)<\/p>\n c)\u00a0(-14\/5, 11\/5)<\/p>\n d)\u00a0(14\/5, 11\/5)<\/p>\n DOWNLOAD APP TO ACESS DIRECTLY ON MOBILE<\/a><\/p>\n Daily Free SSC Practice Set<\/a><\/p>\n Question 11:\u00a0<\/b>If P(x,2) is the centroid of the triangle formed by A(2,y), B(-1,-3) & C(2,6). Then find x*y ?<\/p>\n a)\u00a013<\/p>\n b)\u00a031<\/p>\n c)\u00a03<\/p>\n d)\u00a0-3<\/p>\n Question 12:\u00a0<\/b>What is the reflection of the point (-2,-2) in the line y = -1?<\/p>\n a)\u00a0(-2 , 0)<\/p>\n b)\u00a0(-2, -1)<\/p>\n c)\u00a0(0 , -2)<\/p>\n d)\u00a0(-2, 1)<\/p>\n Question 13:\u00a0<\/b>At what point does the line 3x + 2y = 12 cuts the X-axis?<\/p>\n a)\u00a0(0,6)<\/p>\n b)\u00a0(4,0)<\/p>\n c)\u00a0(-4,0)<\/p>\n d)\u00a0(0,-6)<\/p>\n Question 14:\u00a0<\/b>What is the slope of the line perpendicular to the line 2x+2y=3?<\/p>\n a)\u00a0-2<\/p>\n b)\u00a02<\/p>\n c)\u00a0-1<\/p>\n d)\u00a01<\/p>\n Question 15:\u00a0<\/b>Find the ratio in which P(3,Y) divides the line joining A(1,-3) and B(4,2) ?<\/p>\n a)\u00a01:2<\/p>\n b)\u00a02:1<\/p>\n c)\u00a0-2:1<\/p>\n d)\u00a0-1:2<\/p>\n SSC CGL Free Mock Test<\/a><\/p>\n Download SSC GD Constable Syllabus PDF<\/a><\/p>\n Question 16:\u00a0<\/b>Point P is the midpoint of segment AB. Coordinates of P are (2 , -1) and B are (5, -4) What are the coordinates of point A?<\/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 17:\u00a0<\/b>What is the reflection of the point (-1 , 1) in the line y = 0?<\/p>\n a)\u00a0(-2, 1)<\/p>\n b)\u00a0(-2, -1)<\/p>\n c)\u00a0(-1, -1)<\/p>\n d)\u00a0(1, -1)<\/p>\n Question 18:\u00a0<\/b>What is the slope of the line perpendicular to the line passing through the points (-3 , 1) and (-2 , 2)?<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a0-1<\/p>\n d)\u00a02<\/p>\n Question 19:\u00a0<\/b>If P(1,2) is the centroid of the triangle formed by A(2,3), B(-1,-3) & C(x,y). Then find x-y ?<\/p>\n a)\u00a0-4<\/p>\n b)\u00a04<\/p>\n c)\u00a03<\/p>\n d)\u00a0-3<\/p>\n Question 20:\u00a0<\/b>In what ratio does the point T(1, y) divide the segment joining the points S(3, -1) and U(-1, 2)?<\/p>\n a)\u00a02 : 1<\/p>\n b)\u00a01 : 3<\/p>\n c)\u00a03 : 2<\/p>\n d)\u00a01 : 1<\/p>\n Question 21:\u00a0<\/b>What is the reflection of the point (4 , 3) in the line y = -1?<\/p>\n a)\u00a0(4, 5)<\/p>\n b)\u00a0(-4, 5)<\/p>\n c)\u00a0(4, -5)<\/p>\n d)\u00a0(-4, -5)<\/p>\n Question 22:\u00a0<\/b>Find the length of the median BE of a triangle ABC, with vertices A(1,2) , B(2,3) and C(3,6) ?<\/p>\n a)\u00a01 unit<\/p>\n b)\u00a02 units<\/p>\n c)\u00a03 units<\/p>\n d)\u00a04 units<\/p>\n Question 23:\u00a0<\/b>Find the slope of the line joining A(-1,2) and B(1, -2) ?<\/p>\n a)\u00a0-2<\/p>\n b)\u00a00<\/p>\n c)\u00a02<\/p>\n d)\u00a0undefined<\/p>\n Question 24:\u00a0<\/b>If P(2, 3) is the midpoint of A(4, y) and B(x, 1), then find x ?<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a00<\/p>\n d)\u00a0-1<\/p>\n Question 25:\u00a0<\/b>What is the reflection of the point ( 2, -2) in the line y = -2?<\/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 26:\u00a0<\/b>Find the equation of median BE of the triangle ABC, with vertices A(1,2) , B(2,3) and C(3,6) ?<\/p>\n a)\u00a0x=2<\/p>\n b)\u00a0x=3<\/p>\n c)\u00a0y=2<\/p>\n d)\u00a0y=3<\/p>\n 100+ Free GK Tests for SSC Exams<\/a><\/p>\n Download Free GK PDF<\/a><\/p>\n Question 27:\u00a0<\/b>What is the slope of the line perpendicular to the line 4x-3y=5 ?<\/p>\n a)\u00a0-3\/4<\/p>\n b)\u00a04\/3<\/p>\n c)\u00a03\/4<\/p>\n d)\u00a0-4\/3<\/p>\n Question 28:\u00a0<\/b>If P(2, 3) is the midpoint of A(4, y) and B(x, 1), then find y ?<\/p>\n a)\u00a02<\/p>\n b)\u00a03<\/p>\n c)\u00a04<\/p>\n d)\u00a05<\/p>\n Question 29:\u00a0<\/b>What is the reflection of the point ( 5, 2) in the line y = -1?<\/p>\n a)\u00a0(-5, 4)<\/p>\n b)\u00a0(-5, -4)<\/p>\n c)\u00a0(5, -4)<\/p>\n d)\u00a0(5, 4)<\/p>\n Question 30:\u00a0<\/b>Find the slope of the line joining A(-3, 2) and B(1, 4) ?<\/p>\n a)\u00a0-1\/2<\/p>\n b)\u00a02<\/p>\n c)\u00a01\/2<\/p>\n d)\u00a0-1<\/p>\n Question 31:\u00a0<\/b>What is the reflection of the point ( 1, 2) in the line x = -3?<\/p>\n a)\u00a0(-7, -2)<\/p>\n b)\u00a0(-7, 2)<\/p>\n c)\u00a0(7, -2)<\/p>\n d)\u00a0(7, 2)<\/p>\n Question 32:\u00a0<\/b>What is the reflection of the point ( 1, 2) in the line x = -3?<\/p>\n a)\u00a0(-7, -2)<\/p>\n b)\u00a0(-7, 2)<\/p>\n c)\u00a0(7, -2)<\/p>\n d)\u00a0(7, 2)<\/p>\n Question 33:\u00a0<\/b>Find the length of the median AD of a triangle ABC, with vertices A(1,2) , B(2,3) and C(0,5) ?<\/p>\n a)\u00a01 unit<\/p>\n b)\u00a02 units<\/p>\n c)\u00a03 units<\/p>\n d)\u00a04 units<\/p>\n Question 34:\u00a0<\/b>Find the slope of the line joining A(1,2) and B(1, -2) ?<\/p>\n a)\u00a0-4<\/p>\n b)\u00a00<\/p>\n c)\u00a02<\/p>\n d)\u00a0undefined<\/p>\n Question 35:\u00a0<\/b>Find the mid-point of the line joining A(3, -2) and B(-3, -2) ?<\/p>\n a)\u00a0(3, -2)<\/p>\n b)\u00a0(-3, 0)<\/p>\n c)\u00a0(0, -2)<\/p>\n d)\u00a0(0, 2)<\/p>\n Question 36:\u00a0<\/b>What is the reflection of the point ( -3, 2) in the line x = -2?<\/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 37:\u00a0<\/b>Find the slope of the line joining A(-1, -5) and B(2, 4)<\/p>\n a)\u00a0-3<\/p>\n b)\u00a00<\/p>\n c)\u00a03<\/p>\n d)\u00a0-1<\/p>\n Question 38:\u00a0<\/b>Find the centroid of the triangle with vertices A(2,3), B(-3,2) & C(1,4) ?<\/p>\n a)\u00a0(0,0)<\/p>\n b)\u00a0(2,-3)<\/p>\n c)\u00a0(0,3)<\/p>\n d)\u00a0(3,0)<\/p>\n Question 39:\u00a0<\/b>What is the reflection of the point (4,2) about origin ?<\/p>\n a)\u00a0(-4, -2)<\/p>\n b)\u00a0(4, 2)<\/p>\n c)\u00a0(-4, 2)<\/p>\n d)\u00a0(4, -2)<\/p>\n Question 40:\u00a0<\/b>: Find the equation of the median AD of a triangle ABC, with vertices A(1,2) , B(2,3) and C(3,1) ?<\/p>\n a)\u00a0y=1<\/p>\n b)\u00a0y=2<\/p>\n c)\u00a0y=3<\/p>\n d)\u00a0y=4<\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\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 A(2, 3) is the orthocentre.<\/p>\n So the answer is option D.<\/p>\n 2)\u00a0Answer\u00a0(B)<\/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, $(x-1)^2+(y-2)^2=4^2$<\/p>\n ==> $x^2-2x+1+y^2-4y+4 = 16$<\/p>\n ==> $x^2+y^2-2x-4y-11 = 0$<\/p>\n So the answer is option B.<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n (-4, 2) lies on left to the line x=-3<\/p>\n Distance between the line & point is -3-(-4) = 1<\/p>\n So its reflection must be 1 unit to the right side the line , i.e; (-3+1, 2) = (-2, 2)<\/p>\n So the answer is option B.<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Slope = $\\frac{y_2-y_1}{x_2-x_1}$ = $\\frac{2+1}{2-5} = 3\/-3 = -1$<\/p>\n So the answer is option B.<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let (x, y) be the coordinates of A, then<\/p>\n $(\\frac{x-5}{2} , \\frac{y-4}{2}) = (-1, -1)$<\/p>\n x-5 = -2 & y-4 = -2<\/p>\n x = 3 & y = 2<\/p>\n A = (x, y) = (3,2)<\/p>\n So the answer is option C.<\/p>\n SSC GD FREE STUDY MATERIAL<\/a><\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $P(x, y) = (\\frac{mx_2+nx_1}{m+n}, \\frac{my_2+ny_1}{m+n})$<\/p>\n $P(x, 3) = (\\frac{-1m+3n}{m+n}, \\frac{2m-n}{m+n})$<\/p>\n $\\frac{2m-n}{m+n} = 3$<\/p>\n $2m-n = 3m+3n$<\/p>\n $-m = 4n$<\/p>\n $m\/n = -4\/1$<\/p>\n So the answer is option B.<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n (-4, 3) lies 3 units left to the line x = -1<\/p>\n So its reflection must lie 3 unit right to the line,<\/p>\n So its reflection is (-1+3, 3) = (2, 3)<\/p>\n So the answer is option D.<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 3x – 2y = 12 lines cuts Y-axis, where x = 0<\/p>\n 3(0)-2y = 12<\/p>\n -2y = 12<\/p>\n y = -6<\/p>\n So the point is (0, -6)<\/p>\n So the answer is option A.<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Slope of a line ax+by+c = 0 is -a\/b<\/p>\n Slope of line x-3y+2=0 = -1\/-3 = 1\/3<\/p>\n Slope of parallel line = 1\/3<\/p>\n So the answer is option B.<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $P(x, y) = (\\frac{mx_2+nx_1}{m+n}, \\frac{my_2+ny_1}{m+n})$<\/p>\n $P(x, y) = (\\frac{2(4)+3(2)}{2+3}, \\frac{2(1)+3(3)}{2+3})$<\/p>\n $P(x, y) = (\\frac{14}{5}, \\frac{11}{5})$<\/p>\n So the answer is option D.<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Centroid $= (\\frac{x_1+x_2+x_3}{3}, \\frac{y_1+y_2+y_3}{3})$<\/p>\n $(x, 2) = (\\frac{2-1+2}{3}, \\frac{y-3+6}{3})$<\/p>\n $(x, 2) = (1, \\frac{y+3}{3})$<\/p>\n => $x = 1$ , & $2 = (y+3)\/3$<\/p>\n => $x = 1$, & $y = 3$<\/p>\n $x*y = 3$<\/p>\n So the answer is option C.<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n (-2, -2) lies 1 unit below the line y = -1<\/p>\n So its reflection must lie 1 unit above the line,<\/p>\n So its reflection is (-2, -1+1) = (-2, 0)<\/p>\n So the answer is option A.<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Put y = 0 in 3x + 2y = 12,<\/p>\n 3x = 12<\/p>\n x = 4<\/p>\n So the point is (4,0)<\/p>\n So the answer is option B.<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n slope of the line perpendicular to the line ax+by+c=0 is, b\/a<\/p>\n slope of the line perpendicular to the line 2x+2y=3 is 2\/2=1<\/p>\n So the answer is option D.<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the ratio be m:n<\/p>\n (3, Y) = ($\\frac{mx_2+nx_1}{m+n},\\frac{my_2+ny_1}{m+n}$)<\/p>\n 3 = $\\frac{4m+n}{m+n}$<\/p>\n 3m+3n = 4m+n<\/p>\n 2n = m<\/p>\n m\/n = 2\/1<\/p>\n So the answer is option B.<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let A be (x, y)<\/p>\n (2,-1) = ($\\frac{x+5}{2} , \\frac{y-4}{2}$)<\/p>\n x+5 = 2*2 and y-4 = -1*2<\/p>\n x = 4-5 and y = -2+4<\/p>\n x = -1 and y = 2<\/p>\n A = (x,y) = (-1,2)<\/p>\n So the answer is option B.<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n From the diagram answer is (-1,-1)<\/p>\n So the answer is option C.<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Slope = m = $\\frac{y_2-y_1}{x_2-x_1}$ = $\\frac{2-1}{-2+3}$ = $1$<\/p>\n Slope of perpendicular line = -1\/m = -1\/1 = -1<\/p>\n So the answer is option C.<\/p>\n 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n ($\\frac{2-1+x}{3} , \\frac{3-3+y}{3}$) = (1, 2)<\/p>\n ($\\frac{1+x}{3} , \\frac{y}{3}$) = (1, 2)<\/p>\n 1+x = 3 and y = 6<\/p>\n x = 2 and y = 6<\/p>\n x-y = 2-6 = -4<\/p>\n So the answer is option A.<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the ratio is m:n<\/p>\n 1 = $\\frac{m(-1)+n(3)}{m+n}$<\/p>\n $m+n = -m+3n$<\/p>\n $2m = 2n$<\/p>\n $m:n = 1:1$<\/p>\n So the answer is option D.<\/p>\n 21)\u00a0Answer\u00a0(C)<\/strong><\/p>\n (4, 3) lies above the line y=-1<\/p>\n Distance b\/w the line & point is 3-(-1) = 4<\/p>\n So its reflection must be 4 units below the line , i.e; (4, -1-4) = (4, -5)<\/p>\n So the answer is option C.<\/p>\n 22)\u00a0Answer\u00a0(A)<\/strong><\/p>\n E = mid point of AC = $(\\frac{x_{1}+x_{2}}{2} , \\frac{y_{1}+y_{2}}{2})$ = $(\\frac{1+3}{2} , \\frac{2+6}{2})$ = $(2, 4)$<\/p>\n B = (2 , 3) & E = ( 2 , 4)<\/p>\n Since x values are same,distance = $y_{2}-y_{1} = 4-3 = 1$<\/p>\n So the answer is option A.<\/p>\n 23)\u00a0Answer\u00a0(A)<\/strong><\/p>\n slope m = $\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{-2-2}{1+1}=-4\/2=-2$<\/p>\n So the answer is option A.<\/p>\n 24)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Mid point = $(\\frac{x_{1}+x_{2}}{2} , \\frac{y_{1}+y_{2}}{2})$<\/p>\n (2, 3) = $(\\frac{4+x}{2} , \\frac{1+y}{2})$<\/p>\n $\\frac{4+x}{2}$ = 2<\/p>\n 4+x = 4<\/p>\n x = 0<\/p>\n So the answer is option C.<\/p>\n 25)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The point (2, -2) lies on the line y = -2, so reflection is itself<\/p>\n So the answer is option A.<\/p>\n 26)\u00a0Answer\u00a0(A)<\/strong><\/p>\n E = mid point of AC = $(\\frac{x_{1}+x_{2}}{2} , \\frac{y_{1}+y_{2}}{2})$ = $(\\frac{1+3}{2} , \\frac{2+6}{2})$ = $(2, 4)$<\/p>\n B = (2 , 3) & E = ( 2 , 4)<\/p>\n Since x values are same, Equation of BE is x = 2<\/p>\n So the answer is option A.<\/p>\n 27)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Slope of a line ax+by+c = 0 is $\\frac{-a}{b}$<\/p>\n Slope of the line 4x-3y=5 is m =-4\/-3 = 4\/3<\/p>\n Slope of a line perpendicular to this line = -1\/m = -3\/4<\/p>\n So the answer is option A.<\/p>\n 28)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Mid point = $(\\frac{x_{1}+x_{2}}{2} , \\frac{y_{1}+y_{2}}{2})$<\/p>\n (2, 3) = $(\\frac{4+x}{2} , \\frac{1+y}{2})$<\/p>\n $\\frac{1+y}{2}$ = 3<\/p>\n 1+y = 6<\/p>\n y = 5<\/p>\n So the answer is option D.<\/p>\n 29)\u00a0Answer\u00a0(C)<\/strong><\/p>\n From the diagram reflection of (5 , 2) is (5 , -4)<\/p>\n So the answer is option C.<\/p>\n 30)\u00a0Answer\u00a0(C)<\/strong><\/p>\n slope m = $\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{4-2}{1+3}=2\/4=1\/2$<\/p>\n So the answer is option C.<\/p>\n 31)\u00a0Answer\u00a0(B)<\/strong><\/p>\n from the diagram answer is (-7 , 2)<\/p>\n So the answer is option B.<\/p>\n 32)\u00a0Answer\u00a0(B)<\/strong><\/p>\n from the diagram answer is (-7 , 2)<\/p>\n So the answer is option B.<\/p>\n 33)\u00a0Answer\u00a0(B)<\/strong><\/p>\n D = mid point of BC = $(\\frac{x_{1}+x_{2}}{2} , \\frac{y_{1}+y_{2}}{2})$ = $(\\frac{2+0}{2} , \\frac{3+5}{2})$ = $(1, 4)$<\/p>\n A = (1 , 2) & D = ( 1 , 4)<\/p>\n Since x values are same, Distance = $y_{2}-y_{1}=4-2=2$<\/p>\n So the answer is option B.<\/p>\n 34)\u00a0Answer\u00a0(D)<\/strong><\/p>\n slope m = $\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{-2-2}{1-1}=\\infty$<\/p>\n So the answer is option D.<\/p>\n 35)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Mid-point = $(\\frac{x_{1}+x_{2}}{2} , \\frac{y_{1}+y_{2}}{2})$ = $(\\frac{3-3}{2} , \\frac{-2-2}{2})$ = $( 0 , -2)$<\/p>\n So the answer is option C.<\/p>\n 36)\u00a0Answer\u00a0(B)<\/strong><\/p>\n from the diagram answer is (-1 , 2)<\/p>\n So the answer is option B.<\/p>\n 37)\u00a0Answer\u00a0(C)<\/strong><\/p>\n slope m = $\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{4+5}{2+1}=9\/3=3$<\/p>\n So the answer is option C.<\/p>\n 38)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Centroid = $(\\frac{x_{1}+x_{2}+x_{3}}{3} , \\frac{y_{1}+y_{2}+y_{3}}{3})$ = $(\\frac{2-3+1}{3} , \\frac{3+2+4}{3})$ = $(\\frac{0}{3} , \\frac{9}{3})$ = (0, 3)<\/p>\n So the answer is option C.<\/p>\n 39)\u00a0Answer\u00a0(A)<\/strong><\/p>\n To find reflection about origin, just change the signs.<\/p>\n reflection of the point (4,2) about origin is (-4, -2)<\/p>\n So the answer is option A.<\/p>\n 40)\u00a0Answer\u00a0(B)<\/strong><\/p>\n A(1,2) , B(2,3) and C(3,1)<\/p>\n D = mid point of B and C = $(\\frac{2+3}{2} , \\frac{3+1}{2})$ = $(\\frac{5}{2} , \\frac{4}{2})$ = $(5\/2 , 2)$<\/p>\n A(1 , 2) , D(5\/2 , 2)<\/p>\n Equation of AD is y = 2 ($\\because$ y value is same for both points)<\/p>\n So the answer is option B.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/113402.PNG\")
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