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17
The equation of line 
L is 
quad y=(1)/(2)x+1 Line 
L and line J
are perpendicular intersect at the point 
(4,3)

3=mx+e
Do not write
outside the
box
Not drawn
accurately
Work out the equation of line 
J.
Give your answer in the form 
y=mx+c
[3 marks]

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Answer 
qquad
6
Turn over

1717\newlineThe equation of line L L is y=12x+1 \quad y=\frac{1}{2} x+1 Line L L and line J\newlineare perpendicular intersect at the point (4,3) (4,3) \newline3=mx+e 3=m x+e \newlineDo not write\newlineoutside the\newlinebox\newlineNot drawn\newlineaccurately\newlineWork out the equation of line J \mathrm{J} .\newlineGive your answer in the form y=mx+c y=m x+c \newline[33 marks]\newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newlineAnswer \qquad \newline66\newlineTurn over

Full solution

Q. 1717\newlineThe equation of line L L is y=12x+1 \quad y=\frac{1}{2} x+1 Line L L and line J\newlineare perpendicular intersect at the point (4,3) (4,3) \newline3=mx+e 3=m x+e \newlineDo not write\newlineoutside the\newlinebox\newlineNot drawn\newlineaccurately\newlineWork out the equation of line J \mathrm{J} .\newlineGive your answer in the form y=mx+c y=m x+c \newline[33 marks]\newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newlineAnswer \qquad \newline66\newlineTurn over
  1. Identify Perpendicular Slope: Since line LL has a slope of 12\frac{1}{2}, the slope of line JJ, which is perpendicular to LL, will be the negative reciprocal of 12\frac{1}{2}. So, the slope of line JJ is 2-2.
  2. Apply Point-Slope Form: We use the point-slope form to find the equation of line J. The point (4,3)(4,3) lies on line J, and we have the slope from the previous step. The point-slope form is yy1=m(xx1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is the point on the line.
  3. Substitute Values: Substitute the slope (2)(-2) and the point (4,3)(4,3) into the point-slope form: y3=2(x4)y - 3 = -2(x - 4).
  4. Simplify Equation: Simplify the equation: y3=2x+8y - 3 = -2x + 8.
  5. Isolate yy: Add 33 to both sides to get yy by itself: y=2x+11y = -2x + 11.

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