17The equation of line L is y=21x+1 Line L and line Jare perpendicular intersect at the point (4,3)3=mx+eDo not writeoutside theboxNot drawnaccuratelyWork out the equation of line J.Give your answer in the form y=mx+c[3 marks]Answer 6Turn over
Q. 17The equation of line L is y=21x+1 Line L and line Jare perpendicular intersect at the point (4,3)3=mx+eDo not writeoutside theboxNot drawnaccuratelyWork out the equation of line J.Give your answer in the form y=mx+c[3 marks]Answer 6Turn over
Identify Perpendicular Slope: Since line L has a slope of 21, the slope of line J, which is perpendicular to L, will be the negative reciprocal of 21. So, the slope of line J is −2.
Apply Point-Slope Form: We use the point-slope form to find the equation of line J. The point (4,3) lies on line J, and we have the slope from the previous step. The point-slope form is y−y1=m(x−x1), where m is the slope and (x1,y1) is the point on the line.
Substitute Values: Substitute the slope (−2) and the point (4,3) into the point-slope form: y−3=−2(x−4).
Simplify Equation: Simplify the equation: y−3=−2x+8.
Isolate y: Add 3 to both sides to get y by itself: y=−2x+11.
More problems from Find equations of tangent lines using limits