The line L has equation 5y + 12x = 298 A circle, C, has centre (7, 9) L is a tangent to C - AQA - A-Level Maths Pure - Question 6 - 2020 - Paper 2

To find the intersection point of the line L given by the equation:

$5y + 12x = 298$

we first rearrange it to the slope-intercept form:

$y = -\frac{12}{5}x + 59.6$

The slope of line L is $m_L = -\frac{12}{5}$. The gradient of the radius to the tangent point will be the negative reciprocal, which is:

$m_{radius} = \frac{5}{12}$

Using the point (7, 9) as the center of the circle C, we can write the equation of the radius:

$y - 9 = \frac{5}{12}(x - 7)$

Simplifying, we expand this to:

$y - 9 = \frac{5}{12}x - \frac{35}{12}$
$y = \frac{5}{12}x + \frac{81}{12}$

Next, we set the two equations for y equal to each other to find the intersection:

$-\frac{12}{5}x + 59.6 = \frac{5}{12}x + \frac{81}{12}$

Cross-multiplying to eliminate the fractions gives:

$-144x + 3589.2 = 25x + 405$

Solving for x:

$-169x = -3184.2$
$x = 18.86$

Plugging this back into either equation for y, we substitute into the equation of line L:

$y = -\frac{12}{5}(18.86) + 59.6$

Calculating:

$y = 14.12$

Thus, the coordinates of the intersection point are approximately $(19, 14)$ when rounded.