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The equation of the line l is $5 + y - 2x = 0$ - Junior Cycle Mathematics - Question 6 - 2015

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The equation of the line l is $5 + y - 2x = 0$. (a) Find the co-ordinates of the points where l cuts the axes. l cuts the x-axis at (2, 0) and l cuts the y-axis... show full transcript

Worked Solution & Example Answer:The equation of the line l is $5 + y - 2x = 0$ - Junior Cycle Mathematics - Question 6 - 2015

Step 1

Find the co-ordinates of the points where l cuts the axes.

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Answer

To find where line l cuts the axes, we need to determine the x-intercept and y-intercept.

  1. Finding the x-intercept: Set y = 0 in the equation of the line.

    5+02x=05 + 0 - 2x = 0

    Solving for x gives:
    2x=52x = 5
    x=52=2.5x = \frac{5}{2} = 2.5

    So, the line cuts the x-axis at (2.5, 0).

  2. Finding the y-intercept: Set x = 0 in the equation.

    5+y0=05 + y - 0 = 0

    Solving for y gives:
    y=5y = -5

    Thus, the line cuts the y-axis at (0, -5).

Therefore, the co-ordinates are: l cuts the x-axis at (2.5, 0) and l cuts the y-axis at (0, -5).

Step 2

Find the slope of the line l.

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Answer

To find the slope, we can rewrite the equation in the slope-intercept form y=mx+by = mx + b.
Starting from the equation:

5+y2x=05 + y - 2x = 0
Rearranging gives:
y=2x5y = 2x - 5

The slope (m) is the coefficient of x, which is 2. Therefore, the slope of line l is 2.

Step 3

Write down the slope of the line j.

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Answer

Since line j is perpendicular to line l, we take the negative reciprocal of the slope of l.
The slope of line l is 2, so the slope of line j, denoted as mjm_j, is:
mj=12.m_j = -\frac{1}{2}.

Step 4

Find the equation of the line j.

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Answer

We can use the point-slope form of the line equation to find the equation of line j. The point (11, 6) and the slope mj=12m_j = -\frac{1}{2} are given.
The point-slope form is:
yy1=m(xx1)y - y_1 = m(x - x_1)
Substituting the values, we get:
y6=12(x11)y - 6 = -\frac{1}{2}(x - 11)
Simplifying this, we multiply through by -2 to eliminate the fraction:

2(y6)=(x11)-2(y - 6) = (x - 11)

Expanding gives:
2y+12=x11-2y + 12 = x - 11

Rearranging it into standard form results in:
x+2y23=0.x + 2y - 23 = 0.
Therefore, the equation of line j is x + 2y - 23 = 0.

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