Q. 1 ∆𝐴𝐵𝐶 is a right angled at 𝐶. If 𝑝 is the length of the perpendicular from 𝐶 to

𝐴𝐵 and 𝑎, 𝑏, 𝑐 are the lengths of the sides opposite to ∠𝐴, ∠𝐵 and ∠𝐶 respectively, then

prove that $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$.

 

(OR)

 

Q.2 If 𝑎≠𝑏≠0, prove that the points (𝑎, 𝑎²), (b, b²)and (0, 0) will not be collinear.

We are given ∆𝐴𝐵𝐶 is right angled at 𝐶 and the length of perpendicular to 𝐴𝐵 from 𝐶. We need to prove that $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$.

In ∆𝐴𝐶𝐵 and ∆𝐶𝐷𝐵, 

we can say that, ∠𝐴𝐶𝐵 =∠𝐶𝐷𝐵 ∠𝐶𝐵𝐷 is the common angle in both the triangles. Therefore, ∠𝐶𝐵𝐷 =∠𝐶𝐵𝐷 

Hence, by 𝐴𝐴 𝑐𝑟𝑖𝑡𝑒𝑟𝑖𝑜𝑛, ∆𝐴𝐶𝐵~∆𝐶𝐷𝐵, 

So, the ratio of corresponding sides will be equal.

$\begin{array}{l}\Rightarrow \frac{b}{p}==\frac{c}{a}\\ \Rightarrow \frac{1}{p}=\frac{c}{ab}\end{array}$

By squaring both sides, we get

$\frac{1}{p^2}=\frac{c^2}{a^2{b}^2}$

By Pythagoras Theorem we know that the sum of the squares on the legs of a right triangle is equal to the square of the hypotenuse.

$\begin{array}{r}\Rightarrow c^2=a^2+b^2\\ \Rightarrow \frac{1}{p^2}=\frac{a^2+b^2}{a^2{b}^2}\end{array}$

$\begin{array}{r}\Rightarrow \frac{1}{p^2}=\frac{a^2}{a^2{b}^2}+\frac{b^2}{a^2{b}^2}\\ \Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\end{array}$

(OR)

If three points are collinear, then it can be said that the area of the triangle formed by the points is zero.

$\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_1\left(y_1-y_2\right)\right]=0$

On substitution we get

$\Rightarrow \frac{1}{2}\left[\left(a\right)\left(b^2-0\right)+\left(b\right)\left(0-a^2\right)+\left(c\right)\left(a^2-b^2\right)\right]=0$

$\Rightarrow \frac{1}{2}\left[ab\right(b-a\left)\right]=0$

Since a and b are not equal to each other, $\Rightarrow b-a\ne 0$

Therefore, the points $(a,a^2),(b,b^2)$ and (0,0) will not be collinear.