What is the perpendicular line to y=3/4x+12 that passes through (12,3)?
a line that is perpendicular has slope that is the negative reciprocal such that m * m' = -1 the slope-intercept form y = mx + b where m = the slope, and b the y-intercept so for y = (3/4)x + 12 so m = 3/4 => m' = -4/3 so using the point-slope formula y − y₁ = m(x − x₁) => y − 3 = (-4/3)(x − 12) => y − 3 = -4x/3 + 16 => y = -4x/3 + 19
Find the slope of a line that is perpendicular to the line 4x + 3y - 15 = 0.?
4x+3y=15 3y=15-4x y=5-(4/3)x two perpendicular lines their slopes : m1*m2=-1 -4/3*m2=-1 m2=3/4 correct answer is c
What is the equation of a line perpendicular to 3x +4y=16 and passes through the point (-2, 7)?
First, let’s write the equation of your line in the canonical way, that is[math]4y = 16 - 3x[/math][math]y = 4 - \frac{3}{4}x[/math]So we have[math]\frac{dy}{dx} = -\frac{3}{4}[/math]If [math]\theta[/math] is the angle formed by this line and the x-axis, we have[math]\tan \theta = -\frac{3}{4}[/math]And we know that[math]\tan (\theta + \frac{\pi}{2}) = \frac{\sin(\theta + \frac{\pi}{2})}{\cos(\theta + \frac{\pi}{2})} = \frac{\sin(\theta)\cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2})\cos(\theta)}{\cos(\theta)\cos(\frac{\pi}{2}) - \sin(\theta)\sin(\frac{pi}{2})} = \frac{\cos(\theta)}{-\sin(\theta)} = \frac{4}{3}[/math]So we know that the equation of the perpendicular line is of the form[math]y = \frac{4}{3}x + a[/math]Or, to put it in a presentation similar to your initial question[math]3y - 4x = 3a[/math]We know, in particular, that it holds true for [math](-2,7)[/math], so we can write[math]3\times 7 - 4\times (-2) =3a[/math][math]29 = 3a[/math]Your equation is [math]3y[/math][math] -4x = 29[/math]
How can I prove that 3y=4x-14 is perpendicular to 4y=-3x+48?
Let us first convert the equations in the slope intercept form y=mx + c .Here ‘m’ is slope or gradient and ‘c’ is the y intercept.1st equation → 3y =4x -14→ y=4x/3 -14/3So here m1= 4/3 , c1 = -14/3.2nd equation → 4y =-3x +48→ y = -3x/4 +12So here m2 = -3/4 , c2 =12.The condition for two equations to be perpendicular is m1 = -1/m2.So from the slopes of both the equations given we can clearly see that m1 = -1/m2.This is one way to prove the above 2 equations are perpendicular.
Whats the equation of a line that is perpendicular to y = (1/4)x + 5 and contains the point (2, -3).?
Hi Leansquad Johnson, What you want is the equation of a line y = mx + b where i) m is the negative reciprocal of the slope of the line in the question and ii) b is the y-intercept: i) The slope of the line in the question is the number in front of x, which is 1/4. To get m, flip 1/4 upside down and then put a negative sign in front: m = -(4/1) = -4. So far, you have y = -4x + b. ii) To get b, plug (2,-3) into the equation from part i) and solve: -3 = -4*2 + b = -8 + b ==> b = -3 + 8 = 5 The equation you're looking for is y = -4x + 5. Hope that helps! Daniel
What is the gradient of a line perpendicular to the line with equation: y= -0.25x-3?
The known line has the equation y = -0.25x - 3 Since this is in the Slope-Intercept form of y = mx + b, then the slope (m) = -0.25 (only for this line) and the y intercept (b) = -3 (only for this line) Any line that is perpendicular to the above line will have a slope equal to the negative reciprocal of the known slope. Since the known slope = -0.25, then the negative reciprocal of this = -(1/(-0.25)) = 1/0.25 = 4. Therefore, the gradient (slope) of the line perpendicular to y = -0.25x - 3 is 4. The equation for this perpendicular line is y = 4x + b In order to figure out the y intercept for the perpendicular line, we would have to know at least one point on this line. Since we don't know this information, then we leave the equation for the perpendicular line as y = 4x + b. When you have two perpendicular lines, although there is a relationship between the two slopes (negative reciprocal), there is no relationship between the y intercepts.
Perpendicular to the line y=-4x{containing (4,-3)?
first of all, you should know that the slope of a line perpendicular to a certain line is the opposite of the line's reciprocal. in other words, a line's slope multiplied by its perpendicular line's slope is -1. in this case, the slope of your equation's perpendicular line is 1/4 because 1/4 x -4 = -1 . (yes, the -4 was the slope of the line) so now that you know the slope, you need to find out the rest of the equation. to do this, you use the following formula: y-y1=m(x-x1) the y1 and the x1 are the coordinates of the point that the line passes through. in y and x just mean plain y and x. and the m means the slope. so when you plug in the numbers into the formula, it should look like this: y-(-3)=(1/4)(x-4) and here's how you solve it: y+3=(1/4)x-1 y=(1/4)x-4 so that's your final answer. of course, you don't need to write that parenthesis and use the slash form fraction. i just did that because you can't really do the regular fraction form on the computer.
How do I find the equation that is perpendicular to the line 7x - 2y = 8 and intersects the line 7x - 2y = 8 at x = 8?
We first need to get the equation of the desired straight line in the point-slope form: y – y1 = m(x – x1), and then put it in the general form of Ax + By = - C.Since the equation of line that we’re looking for is perpendicular to the given line 7x – 2y = 8, then the slope m2 of the line that we’re looking for is the negative reciprocal of the slope m1 of the given line 7x -2y = 8, i.e., m2 = - (1/m1).Now, finding the slope of the given line 7x - 2y =8 as follows:-2y = -7x + 8, then, solving for y, we get: y = (7/2)x – 4; therefore, the slope of our desired line is m2 = - (1/m1) = - (1/7/2) = -2/7.Since the given line intersects our desired line at x = 8, we know that the two lines have a point in common, and solving for the y-coordinate of the point of intersection, we get:7(8) – 2y = 856 – 2y = 8-2y = -56 + 8-2y = -48y = 24, therefore, substituting, our desired line has the following equation:y – 24 = (-2/7)(x – 8)y – 24 = (-2/7)x + (16/7)7y – 168 = -2x + 162x + 7y = 184 is our desired line.
What is distance from the point (-3,-4) to the line 3x-4y-1=0?
Let us first take look at the general derivation of point to line distance formula.Let (m, n) be the point of intersection of the line ax + by + c = 0 and the line perpendicular to it which passes through the point (x0, y0). The line through these two points is perpendicular to the original line, soThus, and by squaring this equation we obtain:Now consider,using the above squared equation. But we also have,since (m, n) is on ax + by + c = 0. Thus,and we obtain the length of the line segment determined by these two points,so in our casea=3 , b=-4, c=-1x0=-3 and y0=-4therefored=(3*(-3)+(-4)*(-4)+(-1))/sqrt(9+16)d=-(9+16-1)/5d=24/5Distance from a point to a line
Y=-3/4x-6 parallel and perpendicular equations?
Slope-intercept form: y = mx + b, where m is the slope and b is the y-intercept The line y = (-3/4)x - 6 has a slope of -3/4. Parallel lines have the same slope, so the slope of the parallel line is also -3/4. Therefore its equation so far is y = (-3/4)x + b. Now solve for b by substituting the point (9, 4) for x and y into the equation. y = (-3/4)x + b 4 = (-3/4)(9) + b 4 = -27/4 + b 16/4 = -27/4 + b 43/4 = b The equation of the parallel line is y = (-3/4)x + 43/4. Perpendicular lines have slopes that are opposite reciprocals (have a product of -1), so the slope of the perpendicular line is 4/3. Therefore, the equation so far is y = (4/3)x + b. Now solve for b by substituting the point (9, 4) for x and y into the equation. y = (4/3)x + b 4 = (4/3)(9) + b 4 = 36/3 + b 4 = 12 + b -8 = b The equation of the perpendicular line is y = (4/3)x - 8.