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# whats the equation of line perpendicular to the given line, passes through -2+6y=13x+2 and (-2,-4)?

need a written equation for line perpendicular to the given line that passes through the given points

## Research, Knowledge and Information :

### whats the equation of line perpendicular to the given line ...

need a written equation for line perpendicular to the given line that ... whats the equation of line perpendicular to the given line, passes through -2+6y=13x+2 ...

### Write the equation of the line perpendicular to y=-2x+5 and ...

Write the equation of the line perpendicular to y=6 through the point ... whats the equation of line perpendicular to the given line, passes through -2+6y=13x+2 and ...

### Find the Equation of a Line Parallel or Perpendicular to ...

... Parallel or Perpendicular to Another Line ... perpendicular to a given line passing through a specific point. Find the Equation of a Line Parallel or ...

### How to find the equation of a perpendicular line - SAT Math

... How to find the equation of a perpendicular line ... A line passes through (2, 8) and (4, 15). ... The equation of line p is given in the form y = mx + b, ...

### What is the equation of the line perpendicular to x+7y=8, and ...

What is the equation of the line perpendicular to x+ ... of the line perpendicular to the given line, ... the equation for the line that passes through the ...

### what is the equation for the line that is parallel to y=-2/3x ...

what is the equation for the line that is parallel to y=-2/3x+12 and passes through ... given coordinates into your equation y = (-2 ... Equation 2 (Perpendicular ...

### 4%2D4 Parallel and Perpendicular Lines

parallel yx = , 4 = = = = , = = = . and K parallel 2 + 62/87,21 + + = = 12 = = the the ...

### Slopes and Equations of Lines - University of Missouri–St ...

Slopes and Equations of Lines ... Equation of the given perpendicular line and a ... Example 3 What is the equation of a line with slope - 2 which passes through ...

### How to find the equation of a perpendicular line - ACT Math

... How to find the equation of a perpendicular line. ... Plugging in the point given into the equation y = 1/2x + b and ... Line m passes through the points (1, 4) ...

## Suggested Questions And Answer :

### WHATS THE PERPENDICULAR LINE OF -X-3Y=-15;(-2,3)

WHATS THE PERPENDICULAR LINE OF -X-3Y=-15;(-2,3) prependicular to the given line and that passes through the given point -X - 3Y = -15 -3Y = X - 15 Y = -(1/3)X + 5 The slope of this line is -1/3; the slope of a line perpendicular to this one is 3. The equation for the perpendicular line is y = 3x + b To pass through (-2,3), b = y - 3x b = 3 - 3(-2) b = 3 + 6 b = 9 That makes the equation y = 3x + 9

### whats the equation of line perpendicular to the given line, passes through -2+6y=13x+2 and (-2,-4)?

-2 + 6y = 13x + 2 6y = 13x + 4 y = (13/6)x + 2/3 slope = 13/6 perpendicular slope = -6/13 we want the equation of a line with slope = -6/13, passing through point (-2,-4) point-slope form y - y1 = m (x - x1) y - -4 = (-6/13) (x - -2) y + 4 = (-6/13) (x + 2) y + 4 = (-6/13)x - 12/13 y = (-6/13)x - 12/13 -4 y = (-6/13)x -64/13

### whats the equation of line perpendicular to the given line, passes through -2+6y=13x+2 and (-2,-4)?

-2 + 6y = 13x + 2 6y = 13x + 4 y = (13/6)x + 2/3 slope = 13/6 perpendicular slope = -6/13 we want the equation of a line with slope = -6/13, passing through point (-2,-4) point-slope form y - y1 = m (x - x1) y - -4 = (-6/13) (x - -2) y + 4 = (-6/13) (x + 2) y + 4 = (-6/13)x - 12/13 y = (-6/13)x - 12/13 -4 y = (-6/13)x -64/13

### No It's two sentences.

Although you haven't given any line equations, I can tell you how to do this sort of problem. Get the linear equation into the form y=ax+b, where a is the slope and b the y intercept (when x=0). Let the point (X,Y) on the line be the point where the perpendicular passes through it. The equation of the perpendicular will be y=-ax+c, so we have to find c. -a is at right angles to a. We know the perpendicular must pass through (X,Y) so Y=-aX+c and therefore c=Y+aX. Remember X and Y are not variables but actual numbers like a, so c will be a number. The linear equation for the perpendicular is y=-ax+Y+aX.

### how do you find the slope of a line sagment?

how do you find the slope of a line sagment? you are given two points which define a line segment. Find the slope of the line sagment, and then use this slope to write the equation of a parallel line [and a] perpendicular line which pass through the new y-intercept 1. line segment (1,8) and (7,-4)  y-intercept of new lines: (0,-7) The slope is (y2 - y1) / (x2 - x1) m = (-4 - 8) / (7 - 1) = -12/6 = -2 This line segment is y = -2x + b. We don't know the y-intercept, but we are not concerned with that. A parallel line has the same slope but a different y-intercept. We are given the y-intercept, so the equation is y = -2x - 7. A line perpendicular to the given line segment has a slope that is the negative reciprocal: 1/2. The equation of that line is y = (1/2)x - 7.

### Write the equation of a line perpendicular to the given line but passing through the given point. y=-1/2x+1 (4,2)

The gradient of the perpendicular is -1/(-1/2)=2. The equation of the perpendicular is y=2x+b, where b can be found by plugging in the point (4,2): 2=8+b, making b=-6 and y=2x-6.

### Find the equation of the line through (4,3) perpendicular to the line 3x +y = 7.

You can easily write the equation of the line you have in this form: y= -3x +7. Then the condition for a line to be perpendicular to a given line is that its gradient multiplied by the gradient of the given line must give - 1 as result. The given line has gradient 3, then the gradient of the perpendicular line must be - 1/3, since 3 multiplied -1/3 is equal to -1.   To this condition you have to add the passage through the point (4;3) which is: y - 3=-1/3 * (x-3)    which leads to y= -x/3 + 1 and that's the equation of the perpendicular line passing through (4;3)

### which equation represents the line that passes through (2, -4) and is perpendicular to the graph of x-4y=8?

x-4y=8 can be written in standard form: y=x/4-2. The tangent has a slope of 1/4, so the normal (perpendicular) has a slope of -4 and would be represented by the equation y=-4x+b, where we find b by putting in the given point: -4=-8+b, so b=4 and y=-4x+4 (answer d).

### Whats the written equation of line parrallel to given line that passes through -3+17=-9 and (-9,19)?

Assuming that's -3x + 17y = -9... -3x + 17 y = -9 17 y = 3x - 9 y = (3/17)x - 9/17 slope = 3/17 point = (-9, 19) point-slope form: y - y1 = m ( x - x1 ) y - 19 = (3/17) ( x - -9 ) y - 19 = (3/17) ( x + 9 ) y - 19 = (3/17)x + 27/17 y = (3/17)x + 27/17 + 323/17 y = (3/17)x + 350/17

### Find the equation of a straight line

ysin30=-xcos30+2; y=-xtan60+2cosec30=-xtan60+4=-x√3+4. When x=0, y=4 (y-intercept), y=0, x=4√3/3. (i) The slope of the line is -tan60=-√3 and this is also the slope of the parallel line which has the equation: y-3=-tan60(x-4) in slope-intercept form. Therefore y=3-xtan60+4tan60=3+4√3-x√3 or xcos30+ysin30=3sin30+4cos30. For a general point P(x,y) on the line, OX=YP=x and OY=XP=y. OT=OS+ST=OS+QP. OS=xcos30 and QP=ysin30, so OT=xcos30+ysin30=2. OT is the radius of the circle of which the line xcos30+ysin30=2 is a tangent. (ii) ON is the length of the perpendicular on to the parallel line xcos30+ysin30=4cos30+3sin30. By analogy, ON=4cos30+3sin30=2√3+3/2 just as OT=2 for the other line. (iii) The distance between the lines must therefore be the difference in the two radii, ON-OT=2√3+3/2-2=2√3-1/2.