math problem(URGENT)

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Can somebody help me out with this Question?

P is the point (2,3) and Q is the point (9,5)
a) find the equation of the line joining PQ.​
b) find the coordinates of the point where the line PQ intersects the x-axis.​
c) the line y=5 is the line of symmetry of triangle PQR. find the coordinates of R.​
d) find the area of triangle PQR.​
e) calculate the length of PQ and hence calculate the perpendicular distance from R to the line PQ.​
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a) Since you have points P and Q, their equation would be in the form \(y=mx+c\) where

Substituting back into \(y=mx+c\) and substituting x and y coordinates from either P or Q should give \(c=\frac{17}{7}\).

Thus equation of the line joining PQ is \(y=\frac{1}{7}(2x+17)\)

b) At x-axis, y=0 so \(x=-\frac{17}{2}\) and therefore coordinates would be \((-\frac{17}{2},0)\)

c) You'll have to either visualize or sketch this. Since line of symmetry is y=5 and Q has coordinated (9,5), R will have same x-coordinates as P and y-coordinates would be 2(5-3)+3 so R will have coordinates (2,7).

d) We can simply use \(\frac{1}{2}\times\text{base}\times\text{height}\) where base would be the length PR and height would be length mid-of-PR and Q (refer to your sketch). This would give the area \(14 units^{2}\).

e) You can simple use the formula to find length between two points \(\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)

If anything is unclear, feel free to ask and I'll be happy explain in more detail.