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The equation of a line through the intersection of two other lines.

The equation of a line through the intersection of two other lines. There are two methods we can use here: Formula (Derivation on page 294)

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The equation of a line through the intersection of two other lines.

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  1. The equation of a line through the intersection of two other lines. There are two methods we can use here: Formula(Derivation on page 294) If a1x + b1y + c1= 0 and a2x + b2y + c2= 0 are the 2 intersecting lines, then the equation of any line through their intersection is given by: (a1x + b1y + c1) + k(a2x + b2y + c2) = 0 Where k is a constant. Method 2 Use simultaneous equations to find the point of intersection. Then use the two point formula, or find the gradient and then point gradient formula.

  2. Exercise 7.9 Page 296, Q 1→6 Start at Q6 and work towards Q1 Example 1 Find the equation of the line through the point (4, 3) and the intersection of 5x + 3y + 1 = 0 and 2x – 5y –12 = 0. In other words, just write the 2 equations in general form and add the k. (a1x + b1y + c1) + k(a2x + b2y + c2) = 0 a1 = 5, b1 = 3 & c1 = 1 a2 = 2, b2 = –5 & c2 = –12 (5x + 3y + 1) + k(2x– 5y– 12) = 0 As the line passes through the point (2, 6) sub the point into the equation. (5(4) + 3(3) + 1) + k(2(4) – 5(3) – 12) = 0 30 + k(–19) = 0 19k = 30 k = 30/19 (5x + 3y + 1) + 30/19(2x– 5y– 12) = 0 19(5x + 3y + 1) + 30(2x– 5y– 12) = 0 95x + 57y + 19 + 60x– 150y– 360 = 0 155x– 93y– 341 = 0 ÷31 5x– 3y– 11 = 0

  3. Today’s work Exercise 7.9 Page 296 Q 1→6 Start at Q6 and work towards Q1

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