Straight Lines | Class 11 Mathematics Notes

To draw the quadrilateral, plot the points (-4,5), (0,7), (5,-5), and (-4,-2) on the Cartesian plane and join them in order. To find the area, use the shoelace formula:

Vertices in order: A(-4,5), B(0,7), C(5,-5), D(-4,-2)

Area = (1/2) |x1y2 + x2y3 + x3y4 + x4y1 - (y1x2 + y2x3 + y3x4 + y4x1)| = (1/2) |(-4)(7) + 0(-5) + 5(-2) + (-4)(5) - [5(0) + 7(5) + (-5)(-4) + (-2)(-4)]| = (1/2) |(-28) + 0 + (-10) + (-20) - [0 + 35 + 20 + 8]| = (1/2) |(-58) - 63| = (1/2) |-121| = 60.5 units²

So, the area of

Given an equilateral triangle with side length 2a, base lies along y-axis with midpoint at origin.

Base endpoints: Since base length = 2a along y-axis, and midpoint at origin, the base vertices are at (0, a) and (0, -a).

To find the third vertex, let it be at (x, 0) on x-axis (since base is vertical, the apex lies on x-axis).

Distance from apex to each base vertex = side length = 2a.

Using distance formula: Distance between (x,0) and (0,a) = 2a \Rightarrow \sqrt{(x-0)^2 + (0 - a)^2} = 2a \Ri

(i) If PQ is parallel to y-axis, then x-coordinates of P and Q are the same, i.e., x_1 = x_2. Distance = |y_2 - y_1|

(ii) If PQ is parallel to x-axis, then y-coordinates of P and Q are the same, i.e., y_1 = y_2. Distance = |x_2 - x_1|

Let the point on x-axis be (x, 0). Distance from (7,6) to (x,0): \sqrt{(x - 7)^2 + (0 - 6)^2} = \sqrt{(x - 7)^2 + 36} Distance from (3,4) to (x,0): \sqrt{(x - 3)^2 + (0 - 4)^2} = \sqrt{(x - 3)^2 + 16}

Equate distances: \sqrt{(x - 7)^2 + 36} = \sqrt{(x - 3)^2 + 16} Square both sides: (x - 7)^2 + 36 = (x - 3)^2 + 16 Expand: (x^2 - 14x + 49) + 36 = (x^2 - 6x + 9) + 16 Simplify: x^2 - 14x + 85 = x^2 - 6x + 25 Cancel x^2: -14x + 85 = -6x + 25 Bring variables to one side: -14x + 6x = 25 - 85 -8x = -6