Figure 2 shows a right angled triangle LMN. The points L and M have coordinates (−1, 2) and (7, −4) respectively. (a) Find an equation for the straight line passing through the points L and M. Give your answer in the form $ax + by + c = 0$, where a, b and c are integers. Given that the coordinates of point N are (16, p), where p is a constant, and angle LMN = 90°, (b) find the value of p. (c) Given that there is a point K such that the points L, M, N and K form a rectangle, determine the y coordinate of K.

Photo AI

Figure 2 shows a right angled triangle LMN - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 2

Question 9

Figure 2 shows a right angled triangle LMN. The points L and M have coordinates (−1, 2) and (7, −4) respectively. (a) Find an equation for the straight line passin... show full transcript

Worked Solution & Example Answer:Figure 2 shows a right angled triangle LMN - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 2

Step 1

(a) Find an equation for the straight line passing through the points L and M.

96%

114 rated

Answer

To find the equation of the line passing through points L(-1, 2) and M(7, -4), we first calculate the gradient (slope) using the formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Substituting the coordinates:

$m = \frac{-4 - 2}{7 - (-1)} = \frac{-6}{8} = -\frac{3}{4}$

Now, we use the point-slope form of the equation of a line, which is given by:

$y - y_1 = m(x - x_1)$

Choosing point L(-1, 2):

$y - 2 = -\frac{3}{4}(x + 1)$

Expanding this:

$y - 2 = -\frac{3}{4}x - \frac{3}{4}$

Bringing everything to one side:

$\frac{3}{4}x + y + \frac{5}{4} = 0$

Multiplying through by 4 to clear the denominator:

$3x + 4y + 5 = 0$

Therefore, the final equation in the required form is:

$3x + 4y + 5 = 0$

Step 2

(b) find the value of p.

99%

104 rated

Answer

Since triangle LMN is a right-angled triangle where LMN = 90°, we can use the property that the product of the gradients of two perpendicular lines is -1.

First, calculate the gradient of line LM:

From the previous calculation, the gradient of LM (denoted as $m_{LM}$ ) is:

$m_{LM} = -\frac{3}{4}$

Next, we can find the gradient of line MN (denoted as $m_{MN}$ ) using the coordinates of N(16, p):

$m_{MN} = \frac{p - (-4)}{16 - 7} = \frac{p + 4}{9}$

Setting the product of the gradients to -1:

$m_{LM} \cdot m_{MN} = -1$

Substituting in the gradients:

$-\frac{3}{4} \cdot \frac{p + 4}{9} = -1$

Clearing the negatives and multiplying both sides by -36 (to eliminate fractions):

$3(p + 4) = 36$

Dividing by 3:

$p + 4 = 12$

Thus:

$p = 12 - 4 = 8$

So, the value of p is 8.

Step 3

(c) find the y coordinate of K.

96%

101 rated

Answer

Since K needs to form a rectangle with points L, M, and N, the coordinates of K can be determined based on the property that opposite sides must be equal in a rectangle.

Let K have coordinates (x_K, y_K). Since LM is perpendicular to MN, K will share the x-coordinate with M and the y-coordinate with L:

$x_K = 7$ $y_K = 2$

Thus, the coordinates of K are (7, 2). Therefore, the y-coordinate of K is:

$y_K = 2$

Figure 2 shows a right angled triangle LMN - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 2

Worked Solution & Example Answer:Figure 2 shows a right angled triangle LMN - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 2

(a) Find an equation for the straight line passing through the points L and M.

(b) find the value of p.

(c) find the y coordinate of K.

Join the A-Level students using SimpleStudy...