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The lines L₁ and L₂ are parallel - AQA - A-Level Maths Pure - Question 8 - 2022 - Paper 1

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The lines L₁ and L₂ are parallel. L₁ has equation 5x + 3y = 15 and L₂ has equation 5x + 3y = 83. L₁ intersects the y-axis at the point P. The point Q is the point o... show full transcript

Worked Solution & Example Answer:The lines L₁ and L₂ are parallel - AQA - A-Level Maths Pure - Question 8 - 2022 - Paper 1

Step 1

Find the coordinates of Q.

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Answer

To find the coordinates of point Q, we first determine the coordinates of point P.

  1. Determine Point P: The intercept of line L₁ on the y-axis can be found by setting x = 0 in the equation of L₁: 3y = 15 \\ y = 5$$ Thus, point P has coordinates (0, 5).
  2. Equation of PQ: The line segment PQ must be perpendicular to L₂, which has the same gradient as L₁.. The slope of L₁ and L₂ can be determined from their equations: 3y = -5x + 15 \\ y = -\frac{5}{3}x + 5$$ Hence, the slope of L₁ and L₂ is -5/3. The slope of PQ, being perpendicular, is the negative reciprocal: $$\frac{3}{5}$$.
  3. Equation of Line PQ: Using the point-slope form of a line equation: y = \frac{3}{5}x + 5$$
  4. Find Intersection Point Q: Set the equations of L₂ and PQ equal: 5x+3y=835x + 3y = 83 Substitute for y from the PQ equation: 5x + \frac{9}{5}x + 15 = 83 \\ 25x + 9x = 340 \\ 34x = 340 \\ x = 10$$ Substitute x back into the PQ equation: $$y = \frac{3}{5}(10) + 5 = 6$$

Thus, the coordinates of Q are (10, 6).

Step 2

Find a.

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Answer

To find the value of a, we consider the distance from the center of the circle C at (a, -17) to the tangent point on line L₁.

  1. Equation of Lines: The coordinates of the point on L₁ can be calculated using: 5x+3y=15 and P(0,5)5x + 3y = 15\text{ and } P(0, 5).

  2. Midpoint of PQ: Since Q lies on L₂ and is closest to P, we express the midpoint of PQ as: xa=0+102=5, and ya=5+62=5.5x_a = \frac{0 + 10}{2} = 5\text{, and } y_a = \frac{5 + 6}{2} = 5.5 Substitute these into the distance formula for the tangents: d=(5−a)2+(5.5+17)2=rd = \sqrt{(5 - a)^2 + (5.5 + 17)^2} = r, where r is the radius.

  3. Find a: Implement the conditions for the tangents: 5+3(−17)=49⇒20+49=49∑=49, solve for aa=205 + 3(-17) = 49 \Rightarrow 20 + 49 = 49 \\ \sum = 49\text{, solve for } a \\ a = 20

Thus, the value of a is 20.

Step 3

Find the equation of C.

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Answer

Using the value of a found:

  1. Circle Equation: The general form of the equation for a circle is: (x−a)2+(y−b)2=r2(x - a)^2 + (y - b)^2 = r^2 Given a = 20 and b = -17, we thus have: (x−20)2+(y+17)2=r2(x - 20)^2 + (y + 17)^2 = r^2 where r is derived from the distance to either line L₁ or L₂.

  2. Construction of r: If we calculate r using distance from point (20, -17) to line L₁ using distance formula or properties of right triangles, Evaluating yields the equation: (x−20)2+(y+17)2=34(x - 20)^2 + (y + 17)^2 = 34.

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