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The line joining the points A(-1,3) and B(5,15) meets the axes at P and Q. Find the equation of AB and calculate the length of PQ. How to calculate the length of PQ??

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    Well,... do that. Where are you having trouble. – 2017-02-24
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    How to calculate the length of PQ – 2017-02-24
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    Collinearity and the fact that each of $P$ and $Q$ already has one determined coordinate. Afterwards Pythagoras' Theorem. – 2017-02-24
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    Well first determine the equation of AB using two point form. – 2017-02-24
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    Ive already done that but ive got no idea how to calculate the length of PQ bcs i dont rlly get what the question is – 2017-02-24

2 Answers 2

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General equation of a line: $y = mx + c$ with $m,c$ being constants $\in \mathbb{R}$.

$m$ is given by $m = \frac{y_{1} - y_{2}}{x_{1}-x_{2}}$ where $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are two distinct points on the line.

Using this we simply plug in the values:

$m = \frac{15-3}{5--1} = 2$

So now we have:

$y = 2x+c$

To find c, we plug in a point on the line into our equation:

$3 = 2*(-1) + c \Rightarrow c = 5$

So our equation is:

$y = 2x + 5$

It meets the axes when $x=0$ and $y=0$. Then without loss of generality let P lie on the x-axis, and Q lie on the y-axis.

$P_{y} = 0 \Rightarrow P_{x} = -2.5$

$Q_{x} = 0 \Rightarrow Q_{y} = 5$

Using Pythagoras's theorem, $PQ = \sqrt{2.5^{2} + 5^{2}}$ = $\frac{5\sqrt{5}}{2}$

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Sketch diagram and dimensions. By comparing dimensions and using similar triangles, it is clear that $OP=5, OQ=\frac 52$ where $O$ is the origin. Hence $$\color{red}{PQ=\frac 52\sqrt{5}}$$