The straight line $l_1$, shown in Figure 1, has equation $5y = 4x + 10$ - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1
Question 10
The straight line $l_1$, shown in Figure 1, has equation $5y = 4x + 10$.
The point $P$ with $x$ coordinate $5$ lies on $l_1$.
The straight line $l_2$ is perpendic... show full transcript
Worked Solution & Example Answer:The straight line $l_1$, shown in Figure 1, has equation $5y = 4x + 10$ - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1
Step 1
Find an equation for $l_2$, writing your answer in the form $ax + by + c = 0$ where $a$, $b$, and $c$ are integers.
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Answer
To find the equation of line l2, we start by determining the gradient of line l1 from its equation 5y=4x+10.
Determine the Gradient of l1:
Rearranging gives:
y=54x+2
Thus, the gradient of l1 is 54.
Find the Gradient of l2:
Since l2 is perpendicular to l1, its gradient will be the negative reciprocal:
ml2=−45
Coordinates of Point P:
When x=5, substituting into the equation of l1 gives:
5y=4(5)+10⇒5y=30⇒y=6
Hence, point P is (5,6).
Forming the Equation of l2:
Using the point-slope form y−y1=m(x−x1):
y−6=−45(x−5)
Expanding this:
y−6=−45x+425
Multiplying through by 4 to eliminate the fraction:
4y−24=−5x+25
Rearranging gives:
5x+4y−49=0
This is the required equation of line l2.
Step 2
Calculate the area of triangle $SPT$.
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Answer
To calculate the area of triangle SPT, we first need the coordinates of points S and T.
Finding Point S:
For line l1, set y=0 to find S:
5(0)=4x+10⇒4x=−10⇒x=−25
So, S is at ig(-\frac{5}{2}, 0\big).
Finding Point T:
For line l2, set y=0 to find T:
0=−45x+425⇒45x=425⇒x=5
So, T is at (5,0).
Coordinates of Points:
S(−25,0)
P(5,6)
T(5,0)
Using the Area Formula for a Triangle:
The area A of triangle formed by points (x1,y1), (x2,y2), and (x3,y3) is given by:
A=21×∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣
Substituting the coordinates:
A=21×−25(6−0)+5(0−0)+5(0−6)A=21×∣−15∣=215
Therefore, the area of triangle SPT is 215 square units.
