Question

The points A(1,-2),B(2,3),C(-3,2) and D(-4,-3) are the vertices of the parallelogram ABCD . Find the altitude of the parallelogram corresponding to the base AB.

Answer 1

i think its help you

pls mark me brainliest answer

ABCD is a parallelogram
A (1,-2), B (2,3), C(-3.2), D(-4,-3)
AB is the base
Next we find out the equation of line
AB.
According to two points formula,
equation of a line passing through two
points [math](x_1, y_1)[/math] and
[math](x_2, y_2)[/math] is
[math]\ y-y_2 = \frac{y_1 - y_2}{x_1 -
x_2}(x - x_2)[/math]
Therefore taking the points (1,-2) and
(2,3), equation of AB is
[math]\ y-3 = \frac{(-2-3)}{1-2}(x - 2)[/
math]
[math]\implies y-3 = \frac{-5}{-1}(x-2)[/
math]
[math]\implies y-3 = 5(x-2)[/math]
[math]\implies y-3 = 5x-10[/math]
[math]\implies 5x - y -7 =0 [/math]
From the figure, we can see that the
height of the parallelogram is the
perpendicular distance of point C from
line AB.
Distance (d) of a point
[math](x_1, y_1)[/math] from a line
[math]ax+by+c=0[/math] is given by the
formula
[math] d = \frac{|ax_1 + by_1 + c|}
{\sqrt{a^2 + b^2}} [/math]
Here,
[math](x_1, y_1) is C(-2,3), a=5, b=-1,
c=-7[/math]
Hence height,
[math] h = \frac{|5(-3) -1(2) - 7|}{\sqrt
{5^2 + (-1)^2}}[/math]
[math]\implies h = \frac{|-15 -2 -7|}
{\sqrt{25 +1}}[/math]
[math]\implies h = \frac{|-24|}{\sqrt
{26}}[/math]
[math]\implies h =\frac{24}{\sqrt{26}}[/
math]
Rationalising, we get
[math]\implies h = \frac{24\sqrt{26}}
{26}[/math]
[math]\implies h = \frac{12}{13}\sqrt
{26}[/math]