Questions & Answers

Question

Answers

Answer
Verified

Hint: If three points are collinear then the area of the triangle formed by those three points will be 0.

If $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)$are three points, then the area of triangle formed with these points is

Area of triangle = $\frac{1}{2}\left( {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right)$--- (1)

Given points $(p,q),(m,n)$ and $\left( {p - m,q - n} \right)$ are collinear. Therefore, from equation (1)

The area of triangle = $$\frac{1}{2}\left( {p(n - q + n) + m(q - n - q) + (p - m)(q - n)} \right)$$

$ \Rightarrow 0 = \frac{1}{2}\left( {p(2n - q) - mn + pq - pn - qm + mn} \right)$

$ \Rightarrow 0 = \frac{1}{2}(2pn - pq + pq - pn - qm)$

$ \Rightarrow 0 = pn - qm$

$ \Rightarrow pn = qm$

Hence proved.

Note: Three points A, B and C are said to be collinear means they are lying on the same straight line. Then the area formed by points ABC will be equal to zero. We can also check collinearity by using distance between points that is AB+BC=AC

If $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)$are three points, then the area of triangle formed with these points is

Area of triangle = $\frac{1}{2}\left( {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right)$--- (1)

Given points $(p,q),(m,n)$ and $\left( {p - m,q - n} \right)$ are collinear. Therefore, from equation (1)

The area of triangle = $$\frac{1}{2}\left( {p(n - q + n) + m(q - n - q) + (p - m)(q - n)} \right)$$

$ \Rightarrow 0 = \frac{1}{2}\left( {p(2n - q) - mn + pq - pn - qm + mn} \right)$

$ \Rightarrow 0 = \frac{1}{2}(2pn - pq + pq - pn - qm)$

$ \Rightarrow 0 = pn - qm$

$ \Rightarrow pn = qm$

Hence proved.

Note: Three points A, B and C are said to be collinear means they are lying on the same straight line. Then the area formed by points ABC will be equal to zero. We can also check collinearity by using distance between points that is AB+BC=AC

×

Sorry!, This page is not available for now to bookmark.