What are the co-ordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 ? -Maths 9th

What are the co-ordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 ? -Maths 9th
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Answer :

(d) 2x + 9y + 7 = 0PS being the median of ΔPQR, S is the mid-point of QR, i.e., Coordinates of S ≡ \(\bigg(rac{6+7}{2},rac{-1+3}{2}\bigg)\) = \(\bigg(rac{13}{2},1\bigg)\)Slope of line parallel to PS = Slope of PS= \(rac{1-2}{rac{13}{2}-2}\) = \(rac{-1}{rac{9}{2}}\) = \(rac{-2}{9}.\)∴ Equation of a line parallel to PS passing through the point (1, –1) is(y + 1) = \(rac{-2}{9}\)(x – 1), i.e., 9y + 9 = – 2x + 2 ⇒ 2x + 9y + 7 = 0.
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Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment. -Maths 9th

Description : Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment. -Maths 9th

Last Answer : Steps of Construction (i) Draw a line segment AB = 5.8 cm. (ii) Taking A as centre and radius more than 1/2AB, draw two arcs, one on either side of AB. (iii) Taking B as centre and ... . (iv) Join CD, intersecting AB at point P. Then, line CPD is the required perpendicular bisector of AB.

(i) Find the co-ordinates of the points on the curve xy = 16 at which the normal drawn meet at origin. (ii) Find the points on the curve`4x^(2)+9 y^(2

Description : (i) Find the co-ordinates of the points on the curve xy = 16 at which the normal drawn meet at origin. (ii) Find the points on the curve`4x^(2)+9 y^(2

Last Answer : (i) Find the co-ordinates of the points on the curve xy = 16 at which the normal drawn meet at ... ` at which the normal drawn is parallel to X-axis.

A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. -Maths 9th

Description : A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. -Maths 9th

Last Answer : answer:

WXYZ is a square of side length 30. V is a point on XY and P is a point inside the square with PV perpendicular to XY. PW = PZ = PV – 5. Find PV. -Maths 9th

Description : WXYZ is a square of side length 30. V is a point on XY and P is a point inside the square with PV perpendicular to XY. PW = PZ = PV – 5. Find PV. -Maths 9th

Last Answer : answer: