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Exercise 6.3 · Q10

Q.Find the length of the perpendicular and the co-ordinates of the foot of the perpendicular from (−10,−2)(-10, -2) to the line x+y−2=0x + y - 2 = 0.

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Length: point-to-line distance formula. Foot: parametric relation x−x1a=y−y1b=−(ax1+by1+c)a2+b2\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{-(ax_1+by_1+c)}{a^2+b^2}.

  • Distance =∣−14∣/2=72=|{-14}|/\sqrt2=7\sqrt2; common ratio =7=7, giving foot (−3,5)(-3,5).

We are asked for both the length of the perpendicular from P(−10,−2)P(-10,-2) to the line x+y−2=0x+y-2=0, and the exact point where that perpendicular meets the line (its foot).

Step 1. Identify the line's coefficients. For x+y−2=0x+y-2=0: a=1, b=1, c=−2a=1,\ b=1,\ c=-2. For the point P(−10,−2)P(-10,-2): x1=−10, y1=−2x_1=-10,\ y_1=-2.

Step 2. Compute the length of the perpendicular. Using D=∣ax1+by1+ca2+b2∣D=\left|\dfrac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\right|:

ax1+by1+c=(1)(−10)+(1)(−2)+(−2)=−10−2−2=−14,ax_1+by_1+c = (1)(-10)+(1)(-2)+(-2) = -10-2-2=-14,

D=∣−1412+12∣=142=1422=72.D=\left|\frac{-14}{\sqrt{1^2+1^2}}\right| = \frac{14}{\sqrt2} = \frac{14\sqrt2}{2} = 7\sqrt2.

Step 3. Set up the parametric relation for the foot of the perpendicular. The foot (x,y)(x,y) satisfies

x−x1a=y−y1b=−(ax1+by1+c)a2+b2.\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-(ax_1+by_1+c)}{a^2+b^2}.

Substituting the values from Steps 1–2 (ax1+by1+c=−14ax_1+by_1+c=-14, a2+b2=2a^2+b^2=2): …

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