Question 15 Marks
A point $P (2, -1)$ is equidistant from the points $(a, 7)$ and $(-3, a)$. Find $a.$
Answer
View full question & answer→We know that the distance between the two points $(x1, y1)$ and $(x2, y2)$ is
$d =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
Let the given points be $A = (a, 7)$ and $B = (−3, a)$ and the third point given is $P(2, −1)$.
We first find the distance between P(2, −1) and A =(a, 7) as follows:
$ PA =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$=\sqrt{(a-2)^2+(7-(-1))^2}$
$=\sqrt{(a-2)^2+(7+1)^2}$
$=\sqrt{(a-2)^2+8^2}$
$=\sqrt{(a-2)^2+64}$
Similarly, the distance between $P(2,−1)$ and $B = (−3, a)$ is:
$ PB =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$=\sqrt{(-3-2)^2+(a-(-1))^2}$
$=\sqrt{(-5)^2+(a+1)^2}$
$=\sqrt{25+(a+1)^2}$
Since the point $P(2,−1)$ is equidistant from the points $A(a, 7)$ and $B = (−3, a)$,
$\therefore PA = PB$ that is:
$\rightarrow \sqrt{(a-2)^2+64}=\sqrt{25+(a+1)^2}$
$\rightarrow \left(\sqrt{(a-2)^2+64^2}\right)=\left(\sqrt{25+\left((a+1)^2\right)^2}\right.$
$\Rightarrow (a − 2)^2+ 64 = 25 + (a + 1)^2$
$\Rightarrow (a − 2)^2− (a + 1)^2= 25 − 64$
$\Rightarrow (a^2+ 4 − 4a) − (a^2+ 1 + 2a) = −39 ...( \because (a − b)^2= a^2+ b^2− 2ab, (a + b)^2= a^2+ b^2+ 2ab)$
$\Rightarrow a^2+ 4 − 4a − a^2− 1 − 2a = −39$
$\Rightarrow −6a + 3 = −39$
$\Rightarrow − 6a =−39 −3$
$\Rightarrow −6a = −42$
$\Rightarrow a =\frac{-42}{-6}=7$
Hence, $a = 7.$
$d =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
Let the given points be $A = (a, 7)$ and $B = (−3, a)$ and the third point given is $P(2, −1)$.
We first find the distance between P(2, −1) and A =(a, 7) as follows:
$ PA =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$=\sqrt{(a-2)^2+(7-(-1))^2}$
$=\sqrt{(a-2)^2+(7+1)^2}$
$=\sqrt{(a-2)^2+8^2}$
$=\sqrt{(a-2)^2+64}$
Similarly, the distance between $P(2,−1)$ and $B = (−3, a)$ is:
$ PB =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$=\sqrt{(-3-2)^2+(a-(-1))^2}$
$=\sqrt{(-5)^2+(a+1)^2}$
$=\sqrt{25+(a+1)^2}$
Since the point $P(2,−1)$ is equidistant from the points $A(a, 7)$ and $B = (−3, a)$,
$\therefore PA = PB$ that is:
$\rightarrow \sqrt{(a-2)^2+64}=\sqrt{25+(a+1)^2}$
$\rightarrow \left(\sqrt{(a-2)^2+64^2}\right)=\left(\sqrt{25+\left((a+1)^2\right)^2}\right.$
$\Rightarrow (a − 2)^2+ 64 = 25 + (a + 1)^2$
$\Rightarrow (a − 2)^2− (a + 1)^2= 25 − 64$
$\Rightarrow (a^2+ 4 − 4a) − (a^2+ 1 + 2a) = −39 ...( \because (a − b)^2= a^2+ b^2− 2ab, (a + b)^2= a^2+ b^2+ 2ab)$
$\Rightarrow a^2+ 4 − 4a − a^2− 1 − 2a = −39$
$\Rightarrow −6a + 3 = −39$
$\Rightarrow − 6a =−39 −3$
$\Rightarrow −6a = −42$
$\Rightarrow a =\frac{-42}{-6}=7$
Hence, $a = 7.$