The line $l_1$ has equation $y = -2x + 3$. The line $l_2$ is perpendicular to $l_1$, and passes through the point (5, 6). (a) Find an equation for $l_2$ in the form $\alpha x + by + c = 0$, where $a, b$ and $c$ are integers. The line $l_1$ crosses the x-axis at the point $A$ and the y-axis at the point $B$. (b) Find the x-coordinate of $A$ and the y-coordinate of $B$. Given that $O$ is the origin, (c) find the area of the triangle $OAB$.

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The line $l_1$ has equation $y = -2x + 3$ - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 3

Question 6

The line $l_1$ has equation $y = -2x + 3$. The line $l_2$ is perpendicular to $l_1$, and passes through the point (5, 6). (a) Find an equation for $l_2$ in the for... show full transcript

Worked Solution & Example Answer:The line $l_1$ has equation $y = -2x + 3$ - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 3

Step 1

Find an equation for $l_2$ in the form $\alpha x + by + c = 0$

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Answer

To find the equation of the line $l_2$ , we first need to determine its gradient. Since $l_2$ is perpendicular to $l_1$ , the gradient of $l_2$ can be found by taking the negative reciprocal of the gradient of $l_1$ .

The equation of $l_1$ is $y = -2x + 3$ , which has a gradient of $-2$ . Therefore, the gradient of $l_2$ is: $m_2 = \frac{1}{2}$

Using the point-slope form of the line's equation, we can express $l_2$ passing through the point (5, 6):

y - 6 = \frac{1}{2}(x - 5)

Rearranging this gives:

y - 6 = \frac{1}{2}x - \frac{5}{2}

Multiplying through by 2 to eliminate the fraction:

$2y - 12 = x - 5$

Rearranging into the desired form yields: $x - 2y + 7 = 0$

Step 2

Find the x-coordinate of $A$ and the y-coordinate of $B$

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Answer

To find the x-coordinate of point $A$ , we need to set $y = 0$ in the equation of line $l_1$ :

$0 = -2x + 3$ Solving for $x$ gives:

$2x = 3 \implies x = \frac{3}{2}$

Thus, the x-coordinate of point $A$ is $\frac{3}{2}$ .

Next, to find the y-coordinate of point $B$ , we set $x = 0$ in the equation of line $l_1$ :

$y = -2(0) + 3 = 3$

Thus, the y-coordinate of point $B$ is $3$ .

Step 3

Find the area of the triangle $OAB$

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Answer

To find the area of triangle $OAB$ , we use the formula for the area of a triangle given by: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

In this case, we can take the base as the distance from $O(0, 0)$ to $A(\frac{3}{2}, 0)$ , which is $\frac{3}{2}$ , and the height as the y-coordinate of point $B(0, 3)$ .

Hence, the area is: $\text{Area} = \frac{1}{2} \times \frac{3}{2} \times 3 = \frac{9}{4}$

Therefore, the area of triangle $OAB$ is $\frac{9}{4}$ square units.

The line $l_1$ has equation $y = -2x + 3$ - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 3

Worked Solution & Example Answer:The line $l_1$ has equation $y = -2x + 3$ - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 3

Find an equation for $l_2$ in the form $\alpha x + by + c = 0$

Find the x-coordinate of $A$ and the y-coordinate of $B$

Find the area of the triangle $OAB$

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