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The line $l_1$, shown in Figure 2 has equation $2x + 3y = 26$ - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 1
Question 2
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 f... 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 2 - 2014 - Paper 1
Step 1
Find an equation for the line $l_2$
Answer
To find the equation of line , we first need to determine the slope of line . The equation of line is given by:
Rearranging this into the slope-intercept form gives:
y = -\frac{2}{3}x + \frac{26}{3}$$ Thus, the slope of line $l_1$ is $m_1 = -\frac{2}{3}$. Since line $l_2$ is perpendicular to line $l_1$, its slope $m_2$ can be found using the negative reciprocal: $$m_2 = \frac{3}{2}$$ Since line $l_2$ passes through the origin (0,0), its equation can be written as: $$y = \frac{3}{2}x$$ Therefore, the equation for line $l_2$ is: $$2y = 3x$$ or simply: $$3x - 2y = 0$$Step 2
Find the area of triangle $OBC$
Answer
To find the area of triangle , we first need the coordinates of points and .
Finding point : The -intercept of line can be found by setting in its equation:
3y = 26 \ y = \frac{26}{3}$$ Thus, point $B$ is at (0, $\frac{26}{3}$). **Finding point $C$:** To find the point of intersection $C$ between lines $l_1$ and $l_2$, we set the equations equal. Using the equation of $l_2$: $$y = \frac{3}{2}x$$ after substituting into the equation of $l_1$: $$2x + 3(\frac{3}{2}x) = 26$$ This simplifies to: $$2x + \frac{9}{2}x = 26 \ \frac{4}{2}x + \frac{9}{2}x = 26 \ \frac{13}{2}x = 26 \ x = 4$$ Substituting $x = 4$ back to find $y$: $$y = \frac{3}{2}(4) = 6$$ Therefore, $C$ is at (4, 6). Now with points $O(0,0)$, $B(0, \frac{26}{3})$, and $C(4, 6)$, we can use the formula for the area of a triangle formed by coordinates $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Substituting the points into this formula gives: $$\text{Area} = \frac{1}{2} \left| 0(\frac{26}{3} - 6) + 0(6 - 0) + 4(0 - \frac{26}{3}) \right|$$ This simplifies to: $$\text{Area} = \frac{1}{2} \left| 4(-\frac{26}{3}) \right| = \frac{1}{2} \cdot \frac{104}{3} = \frac{52}{3}$$ Thus, the area of triangle $OBC$ is: $$\text{Area} = \frac{52}{3}$$