Quantitative Aptitude — Coordinate Geometry

To find the distance between two points (x1, y1) and (x2, y2), we use the distance formula: Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2) Given points are (3, 4) and (6, 8). Let (x1, y1) = (3, 4) and (x2, y2) = (6, 8). Distance = sqrt((6 - 3)^2 + (8 - 4)^2) Distance = sqrt((3)^2 + (4)^2) Distance = sqrt(9 + 16) Distance = sqrt(25) Distance = 5 units.

To find the coordinates of a point P(x, y) that divides the line segment joining A(x1, y1) and B(x2, y2) internally in the ratio m:n, we use the section formula: x = (m*x2 + n*x1) / (m + n) y = (m*y2 + n*y1) / (m + n) Given points are A(2, 3) and B(8, 9), and the ratio m:n = 1:2. x = (1*8 + 2*2) / (1 + 2) = (8 + 4) / 3 = 12 / 3 = 4 y = (1*9 + 2*3) / (1 + 2) = (9 + 6) / 3 = 15 / 3 = 5 So, the coordinates of the point are (4, 5).

To find the midpoint M(x, y) of a line segment joining two points (x1, y1) and (x2, y2), we use the midpoint formula: x = (x1 + x2) / 2 y = (y1 + y2) / 2 Given points are P(-5, 7) and Q(3, -1). Let (x1, y1) = (-5, 7) and (x2, y2) = (3, -1). x = (-5 + 3) / 2 = -2 / 2 = -1 y = (7 + (-1)) / 2 = (7 - 1) / 2 = 6 / 2 = 3 So, the midpoint is (-1, 3).

To find the area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3), we use the formula: Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| Given vertices are (1, 2), (4, 2), and (3, 5). Let (x1, y1) = (1, 2) Let (x2, y2) = (4, 2) Let (x3, y3) = (3, 5) Area = 1/2 * |1(2 - 5) + 4(5 - 2) + 3(2 - 2)| Area = 1/2 * |1(-3) + 4(3) + 3(0)| Area = 1/2 * |-3 + 12 + 0| Area = 1/2 * |9| Area = 4.5 square units.

If three points are collinear, the area of the triangle formed by them is 0. Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| = 0 Given points are A(2, 3), B(4, k), and C(6, 7). Let (x1, y1) = (2, 3) Let (x2, y2) = (4, k) Let (x3, y3) = (6, 7) 1/2 * |2(k - 7) + 4(7 - 3) + 6(3 - k)| = 0 |2k - 14 + 4(4) + 18 - 6k| = 0 |2k - 14 + 16 + 18 - 6k| = 0 |-4k + 20| = 0 -4k + 20 = 0 4k = 20 k = 5.

To find the slope (m) of a line passing through two points (x1, y1) and (x2, y2), we use the formula: m = (y2 - y1) / (x2 - x1) Given points are (3, -2) and (-1, 6). Let (x1, y1) = (3, -2) and (x2, y2) = (-1, 6). m = (6 - (-2)) / (-1 - 3) m = (6 + 2) / (-4) m = 8 / (-4) m = -2.

The equation of a line in point-slope form is: y - y1 = m(x - x1) Given point (x1, y1) = (1, 5) and slope m = 3. Substitute the values into the formula: y - 5 = 3(x - 1) y - 5 = 3x - 3 y = 3x - 3 + 5 y = 3x + 2.

To find the perpendicular distance from a point (x0, y0) to a line Ax + By + C = 0, we use the formula: Distance d = |Ax0 + By0 + C| / sqrt(A^2 + B^2) Given point (x0, y0) = (2, 3) and line 3x + 4y - 5 = 0. Here, A = 3, B = 4, C = -5. d = |3(2) + 4(3) - 5| / sqrt(3^2 + 4^2) d = |6 + 12 - 5| / sqrt(9 + 16) d = |13| / sqrt(25) d = 13 / 5 d = 2.6 units.