Question

Using distance formulas show that A (2, 3), B (4, 7) and C (0, -1) are collinear points. Find a point on Y-axis which is equidistant from P (-6, 4) and Q (2, -8).

Answer 1

To solve the first part of the question, we will first find out what are collinear points. Then we will find out what is the distance formula and how it can be used to find the distance between two points whose coordinates are given. We will find the distance AB, BC and AC using the distance formula $\left( \sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} \right)$. After doing this, we will prove BC = AB + AC. To solve the second part, we will assume the point on the y-axis as R (0, y) and then we will equate the distances PR and QR to get the value of y. After doing this, we will get the required point.

Complete step-by-step answer:
We will start by solving the first part. In the first part, we have to prove that the given points are collinear. Three or more points are said to be collinear if they all lie on a straight line. Now, we have to prove that A, B and C are collinear.

If they are collinear, then the sum of the distances of two pairs of points will be equal to the distance of the third pair of points. Now, we will use distance formulas to calculate the distances AB, BC and AC. The distance formula for calculating the distance between $X\left( {{x}_{1}},{{y}_{1}} \right)$ and $Y\left( {{x}_{2}},{{y}_{2}} \right)$ is given by, $XY=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$. Thus, we have,
$AB=\sqrt{{{\left( 4-2 \right)}^{2}}+{{\left( 7-3 \right)}^{2}}}=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 4 \right)}^{2}}}=\sqrt{4+6}=\sqrt{20}$
Similarly, we have,
$BC=\sqrt{{{\left( 4-0 \right)}^{2}}+{{\left( 7-\left( -1 \right) \right)}^{2}}}=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 8 \right)}^{2}}}=\sqrt{16+64}=\sqrt{80}=2\sqrt{20}$
And we also have,
$AC=\sqrt{{{\left( 2-0 \right)}^{2}}+{{\left( 3-\left( -1 \right) \right)}^{2}}}=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 4 \right)}^{2}}}=\sqrt{4+16}=\sqrt{20}$
Thus, we can say that AB + AC = BC. So, it means that A (2, 3), B (4, 7) and C (0, -1) are collinear points.
Now, we will solve the second part. Let the point which is equidistant from both P and Q be R. AS R lies on y-axis, its x-coordinate will be zero. Thus, we have R (0, a). Now, the distance PR and QR are equal. We know that the distance between two points is given by, $XY=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$. Thus, according to the question, we have,
$\begin{aligned} & PR=QR \\\ & \Rightarrow \sqrt{{{\left( 0-\left( -6 \right) \right)}^{2}}+{{\left( a-4 \right)}^{2}}}=\sqrt{{{\left( 0-2 \right)}^{2}}+{{\left( a+8 \right)}^{2}}} \\\ & \Rightarrow \sqrt{{{\left( 6 \right)}^{2}}+{{\left( a-4 \right)}^{2}}}=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( a+8 \right)}^{2}}} \\\ \end{aligned}$
On squaring both sides, we will get,
$\begin{aligned} & {{\left( 6 \right)}^{2}}+{{\left( a-4 \right)}^{2}}={{\left( 2 \right)}^{2}}+{{\left( a+8 \right)}^{2}} \\\ & \Rightarrow 36+{{\left( a-4 \right)}^{2}}=4+{{\left( a+8 \right)}^{2}} \\\ & \Rightarrow 32={{\left( a+8 \right)}^{2}}-{{\left( a-4 \right)}^{2}} \\\ \end{aligned}$
By using identities, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, we get,
$\begin{aligned} & 32=\left( {{a}^{2}}+16a+64 \right)-\left( {{a}^{2}}-8a+16 \right) \\\ & \Rightarrow 32={{a}^{2}}+16a+64-{{a}^{2}}+8a-16 \\\ & \Rightarrow 32=24a+48 \\\ & \Rightarrow 32-48=24a \\\ & \Rightarrow 24a=-16 \\\ & \Rightarrow a=\dfrac{-2}{3} \\\ \end{aligned}$
Hence, the point $R\left( 0,\dfrac{-2}{3} \right)$ is equidistant from both P and Q.

Note: We can also prove the collinearity of three points without using the distance formula. We know that collinear points lie on a line so they can’t form a triangle. Thus, the triangle made by these points will have zero area. Thus,
$Area=\left| \begin{matrix} 2 & 3 & 1 \\\ 4 & 7 & 1 \\\ 0 & -1 & 1 \\\ \end{matrix} \right|$
When we will calculate the area by expanding the determinants, we will get Area = 0, which proves that A, B and C are collinear.