The line $l_1$, shown in Figure 2 has equation $2x + 3y = 26$. The line $l_2$ passes through the origin $O$ and is perpendicular to $l_1$. (a) Find an equation for the line $l_2$. (4) The line $l_1$ intersects the line $l_2$ at the point $C$. Line $l_1$ crosses the $y$-axis at the point $B$ as shown in Figure 2. (b) Find the area of triangle $OBC$. Give your answer in the form $rac{a}{b}$, where $a$ and $b$ are integers to be determined. (6)

A-Level Edexcel Maths Pure Equation of a Straight Line: The line $l

The line $l_1$, shown in Figure 2 has equation $2x + 3y = 26$ - Edexcel - A-Level Maths Pure - Question 10 - 2014 - Paper 1

Question 10

The line $l_1$, shown in Figure 2 has equation $2x + 3y = 26$. The line $l_2$ passes through the origin $O$ and is perpendicular to $l_1$. (a) Find an equation for ... show full transcript

Worked Solution & Example Answer:The line $l_1$, shown in Figure 2 has equation $2x + 3y = 26$ - Edexcel - A-Level Maths Pure - Question 10 - 2014 - Paper 1

Step 1

Find an equation for the line $l_2$

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Answer

To find the equation of line $l_2$ , we first need to determine the slope of line $l_1$ . The equation of line $l_1$ is given by:

$2x + 3y = 26$

Rearranging this, we get:

$3y = -2x + 26$ $y = - rac{2}{3}x + rac{26}{3}$

Thus, the slope of $l_1$ is $- rac{2}{3}$ . Since line $l_2$ is perpendicular to $l_1$ , its slope ( $m_2$ ) is the negative reciprocal:

$m_2 = rac{3}{2}$

Now, as $l_2$ passes through the origin, its equation can be written as:

$y = rac{3}{2}x$

So, the equation for line $l_2$ is:

$l_2: y = rac{3}{2}x$

Step 2

Find the area of triangle $OBC$

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Answer

To find the area of triangle $OBC$ , we first need the coordinates of points $B$ and $C$ .

Finding point $B$ : Since line $l_1$ crosses the $y$ -axis when $x = 0$ :

Substitute $x = 0$ in $2x + 3y = 26$ :

ightarrow y = rac{26}{3}$$

Thus, point $B$ is $(0, rac{26}{3})$ .

Finding point $C$ : To find point $C$ , we solve the equations of lines $l_1$ and $l_2$ together:

From line $l_2$ : $y = rac{3}{2}x$

Substitute this in line $l_1$ :

$2x + 3igg( rac{3}{2}xigg) = 26$

Solving for $x$ :

$2x + rac{9}{2}x = 26$

$rac{13}{2}x = 26$

$x = 4$

Now substituting $x = 4$ back to find $y$ :

$y = rac{3}{2}(4) = 6$

Therefore, point $C$ is $(4, 6)$ .
Calculating the area of triangle $OBC$ : Using the coordinates of points $O(0, 0)$ , $B(0, rac{26}{3})$ , and $C(4, 6)$ :

The formula for the area of a triangle given vertices $(x_1, y_1)$ , $(x_2, y_2)$ , and $(x_3, y_3)$ is:

$ext{Area} = rac{1}{2} ig| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ig|$

Substituting these values:

$ext{Area} = rac{1}{2} ig| 0ig( rac{26}{3} - 6ig) + 0(6 - 0) + 4ig(0 - rac{26}{3}ig) ig|$

$= rac{1}{2} ig| 4 imes - rac{26}{3} ig|$ $= rac{1}{2} imes rac{104}{3} = rac{52}{3}$

Hence, the area of triangle $OBC$ can be expressed as $rac{52}{3}$ with $a = 52$ and $b = 3$ .

The line $l_1$, shown in Figure 2 has equation $2x + 3y = 26$ - Edexcel - A-Level Maths Pure - Question 10 - 2014 - Paper 1

Worked Solution & Example Answer:The line $l_1$, shown in Figure 2 has equation $2x + 3y = 26$ - Edexcel - A-Level Maths Pure - Question 10 - 2014 - Paper 1

Find an equation for the line $l_2$

Find the area of triangle $OBC$

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