What is the equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinates axes whose sum is –1 ? -Maths 9th

What is the equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinates axes whose sum is –1 ? -Maths 9th
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Answer :

Diagonals of a rhombus bisect each other at right angles ⇒ Co-ordinates of mid-points of AC and BD are equal∴ 0 = \(\bigg(rac{4+(-2)}{2},rac{-5+(-1)}{2}\bigg)\) = (1, –3)Slope of BD = \(rac{-5+1}{4+2}\) = \(rac{-4}{6}\) = \(rac{-2}{3}\)∵ AC ⊥ BD, Slope of AC = \(rac{3}{2}\)∴ Equation of AC, passing through O (1, –3) having slope \(rac{3}{2}\) isy + 3 = \(rac{3}{2}\) (x - 1)⇒ 2y + 6 = 3x – 3 ⇒ 2y = 3x – 9.
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Last Answer : Let the co-ordinates of the point of internal division A be (x, y). Then,\(x\) = \(rac{2 imes(-7)+3 imes8}{2+3}\) = \(rac{-14+24}{5}\) = \(rac{10}{5}\) = 2y = \(rac{2 imes4+3 imes9}{2+3}\) = \(rac{8+27}{5}\) = \(rac{35}{5}\) = 7∴ Co-ordinates of the point for internal division are (2, 7).

Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Description : Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Description : Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

Draw the graph of the equation represented by the straight line which is parallel to the x-axis and 3 units above it. -Maths 9th

Description : Draw the graph of the equation represented by the straight line which is parallel to the x-axis and 3 units above it. -Maths 9th

Last Answer : Solution :- Any straight line parallel to x-axis is given by y = a, where a is the distance of the line from the x-axis. Here a = 3. Therefore the equation of the line is y = 3. To draw the graph of this equation plot the points (0, 3) and (3, 3) and join them.

The graph of a linear equation in two variables is always a straight line. True/false -Maths 9th

Description : The graph of a linear equation in two variables is always a straight line. True/false -Maths 9th

Last Answer : answer:

The point at which the two coordinate axes meet is called the -Maths 9th

Description : The point at which the two coordinate axes meet is called the -Maths 9th

Last Answer : (c) The point at which the two coordinate axes meet is called the origin.

The point at which the two coordinate axes meet is called the -Maths 9th

Description : The point at which the two coordinate axes meet is called the -Maths 9th

Last Answer : (c) The point at which the two coordinate axes meet is called the origin.

A straight line passes through the points (5, 0) and (0, 3). The length of the perpendicular from the point (4, 4) on the line is: -Maths 9th

Description : A straight line passes through the points (5, 0) and (0, 3). The length of the perpendicular from the point (4, 4) on the line is: -Maths 9th

Last Answer : (b) \(rac{\sqrt{17}}{2}\)Equation of the line through the points (5, 0) and (0, 3) y - 0 = \(rac{3-0}{0-5}\) (x - 5)⇒ y = \(rac{-3}{5}\)(x - 5)⇒ 5y + 3x - 15 = 0 ∴ Distance of perpendicular from ... (rac{|20+12-15|}{\sqrt{25+9}{}}\) = \(rac{17}{\sqrt{34}}\) units. = \(rac{\sqrt{17}}{2}\) units.

Express the given statement in the form of a linear equation in two variables. The sum of the ordinate and abscissa of a point is 6. -Maths 9th

Description : Express the given statement in the form of a linear equation in two variables. The sum of the ordinate and abscissa of a point is 6. -Maths 9th

Last Answer : Solution :- x+y = 6