Is it easy...................I wonder

Geometry Level 3

Find the sum of all real values of $p$ for which $A(4,7), C(p,3)$ and $B(7,3)$ form a right angled triangle in the Cartesian plane right angled at $C$

The answer is 4.

1 solution

Gautam Jha
Mar 19, 2015

Clearly, $AC^2+CB^2=AB^2$

or, $(p-4)^2 + (-4)^2+(7-p)^2=3^2+(-4)^2$

or, $p^2-11p+28=0$

or, $(p-4)(p-7)=0$

$p=4$ or $p=7$

But $p=7$ makes coordinates of $C(7,3)$ same as those of $B(7,3)$ which is not possible as it will make $A,B$ and $C$ collinear.

So, the only possible value of $p$ is $\boxed{4}$

In Coordinate Geometry , for two perpendicular lines having slopes $m_{1}$ and $m_{2}$ , the following condition holds true : $m_{1} \cdot m_{2} = -1$

So we just have to calculate the slopes of AC and BC and use the condition to get the answer .

$m_{1} = \dfrac{7-3}{4-p} \Rightarrow \frac{4}{4-p} \\ m_{2} = \dfrac{3-3}{7-p} \Rightarrow 0$ $\text{We have}$ $m_1 \cdot m_{2} = -1 \\ \dfrac{7-3}{4-p} \cdot 0 = -1$

Which gives the same result that you got but in a shorter way !

A Former Brilliant Member - 6 years, 2 months ago

Ooh. Good one! Didn'y pay attention to that! XD

Mehul Arora - 6 years, 2 months ago

Hi, can you help me with this please: Exam question

Syed Hamza Khalid - 1 year, 5 months ago

Is it easy...................I wonder

The answer is 4.

1 solution

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