Question

If three points A (h, 0), P (a, b) and B (0, k) lie on a line, show that: $\frac{a}{h}+\frac{b}{k}=1.$

Answer 1

$\text{Given points are A (h, 0), B (a, b) and C (0, k). If A, B and C are collinest, then slope of AB = slope of BC = slope of CA}Let(x_1,y_1)\to A(h,0),(x_2,y_2)\to B(a,b),(x_3,y_3)\to C(0,k)\text{Now, slope of AB = slope of Bc = slope of CA}\therefore \frac{y_2-y_1}{x_2-x_1}=\frac{y_3-y_2}{x_3-x_2}=\frac{y_1-y_3}{x_1-x_3}\frac{b-0}{a-h}=\frac{k-b}{0-a}=\frac{0-k}{h-0}\text{Taking first two terms,}\Rightarrow \frac{b}{a-h}=\frac{k-b}{-a}\Rightarrow -ab=(a-h)(k-b)\Rightarrow -ab=ak-ab-hk+bh\Rightarrow ak+bh=hk\text{Dividing each term by (hk)},\Rightarrow \frac{ak}{hk}+\frac{bh}{hk}=\frac{hk}{hk}\frac{a}{h}+\frac{b}{k}=1\text{Hence proved}$