Place the point N so that DICN would be a parallelogram. Prove that (IN) is perpendicular to (DC).
3º) Prove that the points L, I and N are collinear.​

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

i believe the answer is c.

step-by-step explanation:

the distance (d) between two points (x1,y1) and (x2,y2) is given by the formula

d = √   ((x2-x1)2+(y2-y1)2)

d = √ (-11-6)2+(4-3)2

d = √ )2+(1)2)

d = √ (289+1)

d = √ 290

the distance between the points is 17.0293863659264

the midpoint of two points is given by the formula

midpoint= ((x1+x2)/2,(y1+y2)/2)

find the x value of the midpoint

xm=(x1+x2)/2

xm=(6+-11)/2=-2.5

find the y value of the midpoint

ym=(y1+y2)/2

ym=(3+4)/2=3.5

the midpoint is: (-2.5,3.5)

graphing the two points, midpoint and distance

p1 (6,3)

p2 (-11,4)

midpoint (-2.5,3.5)

the length of the black line is the distance between the points (17.0293863659264)

find the slope of the line connecting the two points

slope =   (y2-y1)   =   (4-3)   =   (1)   = -0.0588235294117647

(x2-x1) (-11-6) (-17)

find the equation of the line passing through the two points

the general equation for a straight line is

y = mx + b

where m represents the slope of the line which we found in the previous step to be -0.0588235294117647

y = -0.06x + b

we substitute x and y for the values from one of our points (6,3)

3 = -.06×6 + b

3 = -0.35 + b

3--0.35 = b

3.35 = b

knowing both b and m, we can contruct the equation of the line

y= -0.06x+ 3.35

x and y intercepts

the x-intercept is a point on the graph where y is zero

using the equation we found in the previous step and substituting zero for y

y= -0.06x+ 3.35

0= -0.06x+ 3.35

0.06x= 3.35

x= 3.35/0.06 = 57.00

the x intercept for this straight line is 57.00

the y-intercept is a point on the graph where x is zero

using the equation we found in the previous step and substituting zero for x

y= -0.06×0+ 3.35

y= 3.35

the y intercept for this straight line is 57.00

the distance (d) between two points (x1,y1) and (x2,y2) is given by the formula

d = √   ((x2-x1)2+(y2-y1)2)

d = √ (-11-6)2+(4-3)2

d = √ )2+(1)2)

d = √ (289+1)

d = √ 290

the distance between the points is 17.0293863659264

the midpoint of two points is given by the formula

midpoint= ((x1+x2)/2,(y1+y2)/2)

find the x value of the midpoint

xm=(x1+x2)/2

xm=(6+-11)/2=-2.5

find the y value of the midpoint

ym=(y1+y2)/2

ym=(3+4)/2=3.5

the midpoint is: (-2.5,3.5)

graphing the two points, midpoint and distance

p1 (6,3)

p2 (-11,4)

midpoint (-2.5,3.5)

the length of the black line is the distance between the points (17.0293863659264)

find the slope of the line connecting the two points

slope =   (y2-y1)   =   (4-3)   =   (1)   = -0.0588235294117647

(x2-x1) (-11-6) (-17)

find the equation of the line passing through the two points

the general equation for a straight line is

y = mx + b

where m represents the slope of the line which we found in the previous step to be -0.0588235294117647

y = -0.06x + b

we substitute x and y for the values from one of our points (6,3)

3 = -.06×6 + b

3 = -0.35 + b

3--0.35 = b

3.35 = b

knowing both b and m, we can contruct the equation of the line

y= -0.06x+ 3.35

x and y intercepts

the x-intercept is a point on the graph where y is zero

using the equation we found in the previous step and substituting zero for y

y= -0.06x+ 3.35

0= -0.06x+ 3.35

0.06x= 3.35

x= 3.35/0.06 = 57.00

the x intercept for this straight line is 57.00

the y-intercept is a point on the graph where x is zero

using the equation we found in the previous step and substituting zero for x

y= -0.06×0+ 3.35

y= 3.35

the y intercept for this straight line is 57.00 and therefore the awnser is c.