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NCERT Exemplar · Q10

Q.If the intercept of a line between the coordinate axes is divided by the point (−5,4)(-5,4) in the ratio 1:21:2, then find the equation of the line.

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Taking the intercept segment from the x-intercept (a,0)(a,0) to the y-intercept (0,b)(0,b), the point (−5,4)(-5,4) divides it in ratio 1:21:2 (from the aa-end), giving a=−152a=-\dfrac{15}{2} and b=12b=12, so the line is 8x−5y+60=08x-5y+60=0.

Setting up

When a line meets the axes, it meets the x-axis at (a,0)(a,0) and the y-axis at (0,b)(0,b). The segment joining these two intercepts is "the intercept between the axes." The point P(−5,4)P(-5,4) divides this segment in the ratio 1:21:2, measured starting from the (a,0)(a,0) end.

Step 1: Apply the section formula

For a point dividing the segment from (x1,y1)(x_1,y_1) to (x2,y2)(x_2,y_2) in ratio m:nm:n:

(mx2+nx1m+n, my2+ny1m+n)\left(\frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n}\right)

Here (x1,y1)=(a,0)(x_1,y_1)=(a,0), (x2,y2)=(0,b)(x_2,y_2)=(0,b), m:n=1:2m:n=1:2.

x-coordinate:

−5=1(0)+2(a)1+2=2a3  ⟹  a=−152-5=\frac{1(0)+2(a)}{1+2}=\frac{2a}{3} \implies a=-\frac{15}{2}

y-coordinate: …

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