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Exercise 6.3 · Q3

Q.Find the distance between the line 4x+3y+4=04x + 3y + 4 = 0 and the point

(i) (−2,4)(-2, 4)
(ii) (7,−3)(7, -3).
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Distance from (x1,y1)(x_1,y_1) to ax+by+c=0ax+by+c=0 is D=∣ax1+by1+ca2+b2∣D=\left|\dfrac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\right|; here a2+b2=16+9=5\sqrt{a^2+b^2}=\sqrt{16+9}=5.

  • (i) ∣4(−2)+3(4)+4∣/5=8/5|4(-2)+3(4)+4|/5 = 8/5.

  • (ii) ∣4(7)+3(−3)+4∣/5=23/5|4(7)+3(-3)+4|/5 = 23/5.

Both parts use the standard perpendicular-distance formula from a point to a line, applied to the same line 4x+3y+4=04x+3y+4=0 but two different points.

Step 1. Set up the distance formula. For the line 4x+3y+4=04x+3y+4=0, a=4, b=3, c=4a=4,\ b=3,\ c=4, so

D=∣4x1+3y1+442+32∣=∣4x1+3y1+45∣.D=\left|\frac{4x_1+3y_1+4}{\sqrt{4^2+3^2}}\right| = \left|\frac{4x_1+3y_1+4}{5}\right|.

Step 2. Part (i): point (−2,4)(-2,4).

D=∣4(−2)+3(4)+45∣=∣−8+12+45∣=∣85∣=85.D=\left|\frac{4(-2)+3(4)+4}{5}\right| = \left|\frac{-8+12+4}{5}\right| = \left|\frac{8}{5}\right| = \frac{8}{5}.

Step 3. Part (ii): point (7,−3)(7,-3).

D=∣4(7)+3(−3)+45∣=∣28−9+45∣=∣235∣=235.D=\left|\frac{4(7)+3(-3)+4}{5}\right| = \left|\frac{28-9+4}{5}\right| = \left|\frac{23}{5}\right| = \frac{23}{5}.

✓Final answer

(i) 85\dfrac{8}{5} units (ii) 235\dfrac{23}{5} units

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