Enroll to XAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>In the xy\u00ad coordinate system, if (a, b) and (a + 3, b + k) are two points on the line defined a)\u00a07\/3<\/p>\n b)\u00a01<\/p>\n c)\u00a09<\/p>\n d)\u00a03<\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n In the xy\u00ad coordinate system, if (a, b) and (a + 3, b + k) are two points on the line defined by the equation x = 3y – 7, then they must satisfy the equation . Question 2:\u00a0<\/b>Area of the triangle formed by the graph of the line 2x – 3y + 6 = 0 along with the coordinate axes is<\/p>\n a)\u00a01\/2 sq. units<\/p>\n b)\u00a03\/2 sq. units<\/p>\n c)\u00a03 sq. units<\/p>\n d)\u00a06 sq. units<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n the line 2x – 3y + 6 = 0 meets the coordinate axes at (-3,0) and (0,2). Question 3:\u00a0<\/b>ABCD is a rectangle in the clockwise direction. The coordinates of A are (1,3) and the coordinates of C are (5,1), the coordinates of vertices B and D satisfy the line $y=2x+c$, then what will be the coordinates of the mid-point of BC.<\/p>\n a)\u00a0$(\\frac{5}{2},\\frac{7}{2})$<\/p>\n b)\u00a0$(\\frac{9}{2},\\frac{5}{2})$<\/p>\n c)\u00a0$(\\frac{9}{5},\\frac{7}{2})$<\/p>\n d)\u00a0$(3,2)$<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> Midpoint of AC passes through the line y = 2x + c<\/p>\n Midpoint of AC = (3,2)<\/p>\n 2 = 6 + c<\/p>\n c = -4<\/p>\n Line passing through B is y = 2x – 4<\/p>\n (slope of AB)(slope of BC) = -1<\/p>\n $\\left(\\ \\frac{\\ y-1}{x-5}\\right)\\left(\\ \\frac{\\ y-3}{x-1}\\right)=-1$<\/p>\n Substituting y = 2x – 4 and solving, we get<\/p>\n B(2,0) or B(4,4)<\/p>\n B cannot lie on X-axis. Therefore, co-ordinates of B = (4,4)<\/p>\n Midpoint of BC = $\\left(\\frac{9}{2},\\frac{5}{2}\\right)$<\/p>\n The answer is option B.<\/p>\n Question 4:\u00a0<\/b>There are 12 points in a two-dimensional plane with following coordinates: Points A, B, C, D, E,F, G have coordinates (1, 0), (2, 0), (3, 0), (4, 0), (5, 0), (6, 0) and (7, 0) respectively. Points H, I, J havecoordinates (1, 1), (2, 2) and (3, 3) respectively. Points K, L have coordinates (4, -2) and (5, -3)respectively. The number of circles possible with these points are?<\/p>\n a)\u00a0179<\/p>\n b)\u00a0158<\/p>\n c)\u00a0168<\/p>\n d)\u00a0147<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n A, B, C, D, E, F, G all lie of a single line<\/p>\n H, I, J all lie on another single line<\/p>\n K, L lie on another single line<\/p>\n If we are given three points which are not collinear than we can draw a circle from these three points<\/p>\n Question 5:\u00a0<\/b>Let T be the triangle formed by the straight line 3x + 5y – 45 = 0 and the coordinate axes. Let the circumcircle of T have radius of length L, measured in the same unit as the coordinate axes. Then, the integer closest to L is<\/p>\n 5)\u00a0Answer:\u00a09<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n In any right triangle, the circumradius is half of the hypotenuse. Here,L=$\\ \\frac{\\ 1}{2}\\ $* the length of the hypotenuse = $\\ \\frac{\\ 1}{2}$($\\sqrt{\\ 15^2+9^2}$) =\u00a0$\\ \\frac{\\ 1}{2}\\ $*$\\sqrt{\\ 306}$ =\u00a0$\\ \\frac{\\ 1}{\\ 2}\\times\\ $17.49 = 8.74<\/p>\n Hence, the integer close to L = 9<\/p>\n Take XAT 2023 Mock Tests<\/a><\/p>\n Question 6:\u00a0<\/b>The diameter of the circumcircle of the triangle formed by the line $24x + 7y =168$ and the coordinate axes is<\/p>\n a)\u00a0$15\\sqrt{2}$<\/p>\n b)\u00a0$24$<\/p>\n c)\u00a0$25$<\/p>\n d)\u00a0$12\\sqrt{3}$<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let’s rewrite the equation $24x + 7y =168$ in intercept form<\/p>\n $\\frac{x}{7}+\\frac{y}{24}$ = 1<\/p>\n $\\therefore$ The co-ordinates of the triangle are (7,0),(0,0),(0,24)<\/p>\n Diameter of the circumcircle formed by the vertices is given by the hypotenuse, which is between (7,0) and (0,24).<\/p>\n Distance between the co-ordinates (7,0) and (0,24) =$\\sqrt{7^{2} +24^{2}}$\u00a0=25.<\/p>\n Hence C is the correct answer.<\/p>\n Question 7:\u00a0<\/b>The coordinates of a triangle ABC are A(1, 5), B(-2, 3), and C(0,-4); find the equation of the median AD? \u00b7<\/p>\n a)\u00a07x-3y+-8= 0<\/p>\n b)\u00a05x-4y+15=O<\/p>\n c)\u00a0x+3y-16=0<\/p>\n d)\u00a011x-4y+9=0<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given that $\\triangle$ABC<\/p>\n Since AD is the median to BC, D will be mid point of BC. So coordinates of D = ($\\frac{-2+0}{2} ,\\frac{3-4}{2}$) =\u00a0(-1 ,$\\frac{-1}{2}$)<\/p>\n Equation of line passing through points A(1,5) and D(-1,$\\frac{-1}{2}$) will be :<\/p>\n $(y-x_{1})=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(Here $(x_{1},y_{1})=(1,5)$ ,\u00a0 $(x_{2},y_{2})=(-1,\\frac{-1}{2})$)<\/p>\n $(y-5)=\\frac{(\\frac{-1}{2})-(5)}{(-1)-(1)}(x-1)$<\/p>\n $4(y-5)=11(x-1)$<\/p>\n $11x-4y+9=0$<\/p>\n Question 8:\u00a0<\/b>ln a plane rectangular coordinate system, points L, M, N and O are represented by the coordinates (-5, 0), (1,-1), (0, 5), and (-1, 5) respectively. Consider a variable point P in the same plane. The minimum value of PL + PM + PN + PO is<\/p>\n a)\u00a0$1+\\sqrt{37}$<\/p>\n b)\u00a0$5\\sqrt{2}+2\\sqrt{10}$<\/p>\n c)\u00a0$\\sqrt{41}+\\sqrt{37}$<\/p>\n d)\u00a0$\\sqrt{41}+1$<\/p>\n e)\u00a0None of these<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(PL + PN)$ will be minimum if P lies on LN, and $(PM + PO)$ will be minimum if P lies on OM.<\/p>\n => P must be the intersection point of the diagonals of the quadrilateral.<\/p>\n $\\therefore$ Min (PL + PM + PN + PO)<\/p>\n = $LN + OM$<\/p>\n = $(\\sqrt{(0 + 5)^2 + (5 – 0)^2}) + (\\sqrt{(1 + 1)^2 + (-1 – 5)^2})$<\/p>\n = $(\\sqrt{25 + 25}) + (\\sqrt{4 + 36})$<\/p>\n = $\\sqrt{50} + \\sqrt{40} = 5 \\sqrt{2} + 2 \\sqrt{10}$<\/p>\n Question 9:\u00a0<\/b>Two diagonals of a parallelogram intersect each other at coordinates (17.5, 23.5). Two adjacent points of the parallelogram are (5.5, 7.5) and (13.5, 16). Find the lengths of the diagonals.<\/p>\n a)\u00a015 and 30<\/p>\n b)\u00a015 and 40<\/p>\n c)\u00a017 and 30<\/p>\n d)\u00a017 and 40<\/p>\n e)\u00a0Multiple solutions are possible<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Using distance formula,<\/p>\n $CX = \\sqrt{(17.5 – 5.5)^2 + (23.5 – 7.5)^2} = \\sqrt{12^2 + 16^2}$<\/p>\n = $\\sqrt{144 + 256} = \\sqrt{400} = 20$<\/p>\n => $AC = 2 \\times CX = 40$<\/p>\n $BX = \\sqrt{(17.5 – 13.5)^2 + (23.5 – 16)^2} = \\sqrt{4^2 + 7.5^2}$<\/p>\n = $\\sqrt{16 + 56.25} = \\sqrt{72.25} = 8.5$<\/p>\n => $BD = 2 \\times BX = 17$<\/p>\n Question 10:\u00a0<\/b>Consider a triangle drawn on the X-Y plane with its three vertices at (41, 0), (0, 41) and (0, 0), each vertex being represented by its (X,Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is<\/p>\n a)\u00a0780<\/p>\n b)\u00a0800<\/p>\n c)\u00a0820<\/p>\n d)\u00a0741<\/p>\n
\nby the equation x = 3y – 7, then k = ?<\/p>\n
\nso , we get a = 3b – 7
\nand a+3 = 3(b+k) – 7
\nOn solving these equations , we get k = 1<\/p>\n
\nSo, base=2 and height = 3
\nArea of the triangle = $\\frac{bh}{2}$
\n= 3<\/p>\n
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